Circles Class 10 Previous Year Questions (Chapter 10)

Circles in CBSE board papers

CBSE asked 170 different Circles questions in the Class 10 Mathematics board papers of 2023, 2024, 2025 and 2026 (main and compartment exams, Standard and Basic courses). 22 of these questions have a worked solution on this page. Most of them test Section 10.3, Number of Tangents from a Point on a Circle (154 of 170).

Where Circles questions came from, 2023–2026

A question printed in several question-paper sets counts once in each table. A question asked in two exams counts in both rows of the first table.

Questions by exam

Year	Exam	Questions asked	Marks	Question types
2026	Main	40	20 × 1 mark · 8 × 2 marks · 12 × 3 marks	19 MCQs, 1 Assertion–Reason and 20 written answers
2025	Main	49	26 × 1 mark · 8 × 2 marks · 15 × 3 marks	23 MCQs, 3 Assertion–Reason and 23 written answers
2025	Compartment	9	3 × 1 mark · 4 × 2 marks · 2 × 3 marks	3 MCQs and 6 written answers
2024	Main	42	18 × 1 mark · 6 × 2 marks · 15 × 3 marks · 1 × 4 marks · 2 × 5 marks	16 MCQs, 2 Assertion–Reason, 1 case study and 23 written answers
2024	Compartment	14	4 × 1 mark · 6 × 2 marks · 4 × 3 marks	3 MCQs, 1 Assertion–Reason and 10 written answers
2023	Main	37	17 × 1 mark · 8 × 2 marks · 9 × 3 marks · 1 × 4 marks · 2 × 5 marks	14 MCQs, 3 Assertion–Reason, 1 case study and 19 written answers
2023	Compartment	6	5 × 1 mark · 1 × 3 marks	4 MCQs, 1 Assertion–Reason and 1 written answer

Questions by NCERT section and marks

NCERT section	1 mark	2 marks	3 marks	4 marks	5 marks	Total
10.2 Tangent to a Circle	7	3	3	1	1	15
10.3 Number of Tangents from a Point on a Circle	79	30	41	1	3	154
Mixed topics	1	0	0	0	0	1
All questions	87	33	44	2	4	170

Most-tested sections

10.3 Number of Tangents from a Point on a Circle: 154 questions
10.2 Tangent to a Circle: 15 questions

From 2023 to 2026, CBSE took most Circles questions from Section 10.3 (154 of 170). These counts describe past papers; they are not predictions.

Topics in Circles

Each NCERT section of the chapter, in book order: what CBSE asked from it and where to practise it.

10.2 Tangent to a Circle

A tangent is a line that intersects the circle at exactly one point, called the point of contact. A tangent is a secant when the two end points of its chord coincide. Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Hence, at any point there is one and only one tangent. A circle has infinitely many tangents, and at most two tangents can be parallel to a given secant.

How to answer questions on this section

Mark the point of contact and the radius through it before writing anything.
For a tangent at a point, join the centre to that point; the radius is perpendicular to the tangent.
In the right triangle formed, use Pythagoras: 12² − 5² = 119, so PQ = √119 cm.
For fill-in-the-blank items, recall: one point, secant, two parallel tangents, point of contact.
For drawing, keep one line tangent and another parallel line a secant to the circle.

Mistakes to avoid in this section

Calling a line that cuts the circle at two points a tangent instead of a secant.
Forgetting that the radius and tangent are perpendicular, so applying Pythagoras wrongly.
Saying more than one tangent can be drawn at the same point of a circle.

CBSE asked 15 questions from this section in the 2023–2026 papers (7 of 1 mark; 3 of 2 marks; 3 of 3 marks; 1 of 4 marks; 1 of 5 marks): 7 MCQs, 1 case study and 7 written answers. 9 were printed in Standard papers and 6 in Basic papers.

Marks	Questions asked	Years	Asked in Standard	Asked in Basic
1 mark	7	2024, 2025, 2026	3	4
2 marks	3	2023, 2024, 2025, 2026	1	2
3 marks	3	2024, 2025, 2026	3	0
4 marks	1	2023	1	0
5 marks	1	2023	1	0

Solved on this page: question 8, question 15, question 16.

10.3 Number of Tangents from a Point on a Circle

A point inside a circle has no tangent through it. A point on the circle has exactly one tangent through it. A point outside the circle has exactly two tangents through it. From an outside point, the two tangent segments have equal length.

How to answer questions on this section

Identify whether the given point is inside, on, or outside the circle first.
Draw radii to points of contact and mark right angles using Theorem 10.1.
Use Theorem 10.2 to set equal tangent lengths from the same external point.
Apply Pythagoras or RHS congruence in the right triangles formed.
For an external point, join it to the centre; this line bisects the angle between tangents.

Mistakes to avoid in this section

Drawing a tangent from a point inside the circle; no such tangent exists.
Taking the distance from the external point to the centre as the tangent length.
Applying equal tangent lengths when the point lies on the circle, not outside it.

CBSE asked 154 questions from this section in the 2023–2026 papers (79 of 1 mark; 30 of 2 marks; 41 of 3 marks; 1 of 4 marks; 3 of 5 marks): 68 MCQs, 11 Assertion–Reason, 1 case study and 74 written answers. 89 were printed in Standard papers and 72 in Basic papers (7 in both).

Marks	Questions asked	Years	Asked in Standard	Asked in Basic
1 mark	79	2023, 2024, 2025, 2026	45	35
2 marks	30	2023, 2024, 2025, 2026	17	14
3 marks	41	2023, 2024, 2025, 2026	26	20
4 marks	1	2024	0	1
5 marks	3	2023, 2024	1	2

A question printed in both courses is counted in both columns (7 in both).

Solved on this page: question 1, question 2, question 3, question 4, question 5, question 6, question 7, question 9, question 10, question 11, question 12, question 13, question 14, question 17, question 18, question 19, question 20, question 21, question 22.

1 more question uses several sections at once; the tables above count it under Mixed topics.

Key ideas used on this page

The tangent at any point of a circle is perpendicular to the radius through the point of contact
Used in questions 1 6 9 10 13 17 18 19
Lengths of tangents drawn from an external point to a circle are equal
Used in questions 4 7 11
Number of Tangents from a Point on a Circle
Used in questions 8 20
Tangent to a Circle
Used in questions 16 21
Tangent is perpendicular to the radius through the point of contact
Used in question 2
Angle between two tangents drawn from an external point
Used in question 3
Perpendicular from the centre to a chord bisects the chord
Used in question 5
Tangent perpendicular to radius through the point of contact
Used in question 12
Distance between parallel tangents to a circle
Used in question 14
Secant of a circle
Used in question 15
Tangent to a circle is perpendicular to the radius through the point of contact
Used in question 22

5 solved questions last asked in 2026

Q1. Two concentric circles are of radii 6 cm and 10 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Mark practised

Marks: 2
Level: Basic
Asked in: 6 paper sets
Years: 2025, 2026
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2026 Mathematics Basic, Set 430/1/1, Q23, 2 marks)
Status: Answer checked step by step

Step-by-step solution

Step 1: Right angle at point of contact
½ mark
Let AB be the chord of the larger circle which touches the smaller circle at P, and let O be the common centre. Join OP. Since AB is a tangent to the smaller circle, OP ⟂ AB.
Why? A tangent to a circle is perpendicular to the radius through the point of contact, so OP is the perpendicular from the centre to the chord AB.
Step 2: Find half the chord
1 mark
In right triangle OAP, AP = √(OA² − OP²) = √(10² − 6²) = √64 = 8 cm.
Why? In a right triangle, Pythagoras theorem gives the half-chord AP from the hypotenuse OA and one leg OP.
Step 3: Length of the chord
½ mark
The perpendicular from the centre to a chord bisects the chord, so AP = PB = 8 cm. Therefore AB = AP + PB = 8 + 8 = 16 cm.
Why? The perpendicular from the centre to a chord bisects the chord, so the two equal halves add up to the whole chord.

Marks: ½ + 1 + ½ = 2 · Split as in the CBSE marking scheme, 2025 · Basic 430/1/1 Q23 (the marks split is taken from the 2025 Basic paper, where this question also appeared)

Answer:

16 cm

Common mistakes to avoid

Giving the half-chord as the final answer, forgetting that the question asks for the whole chord.
Using the difference of the radii as the half-chord instead of applying Pythagoras theorem.
Taking the chord to be the diameter of the larger circle because the circles are concentric, without using the tangency condition.

Why this works

Both circles have the same centre O. The chord AB of the larger circle touches the smaller circle, so AB is a tangent to the smaller circle. By Theorem 10.1, the radius OP to the point of contact is perpendicular to AB. NCERT Example 1 proves that this perpendicular from the centre bisects the chord, so AP = PB. In right triangle OAP, the hypotenuse OA is the radius of the larger circle and the leg OP is the radius of the smaller circle. Therefore AP = √(10² − 6²) = √(100 − 36) = √64 = 8 cm, and AB = AP + PB = 8 + 8 = 16 cm.

Key idea to remember

The tangent at any point of a circle is perpendicular to the radius through the point of contact: see Key ideas used on this page.

Explanation

The chord of the larger circle touches the smaller circle, so it is a tangent to the smaller circle. The radius drawn to the point of contact is therefore perpendicular to the chord. The perpendicular from the centre to a chord bisects the chord, so the chord is made of two equal halves. Pythagoras theorem gives one half, and the two halves add up to the required length.

Important formulas

AP = √(OA² − OP²) · Pythagoras theorem in the right triangle formed by the radius of the larger circle, the radius of the smaller circle, and the half-chord.
AB = AP + PB · The perpendicular from the centre to a chord bisects the chord, so AP = PB.

Q2. If the distance of a tangent to a circle from its centre is 4 cm, then the length of diameter of the circle is :

Mark practised

(A) 2 cm
(B) 4 cm
(C) 8 cm
(D) 16 cm

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025, 2026
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2025 Mathematics Basic, Set 430/1/2, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 8 cm, because the distance of the tangent from the centre is the radius, so diameter = 2 × 4 = 8 cm.
Why? The tangent at any point of a circle is perpendicular to the radius through the point of contact, so this perpendicular distance equals the radius.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/1/2: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 4 cm (B) by taking the given distance as the diameter instead of the radius.
Choosing 16 cm (D) by squaring 4 cm, which has nothing to do with the diameter.
Choosing 2 cm (A) by dividing the distance by 2 instead of doubling it.

Why this works

The distance of a line from a point means the perpendicular distance. For a tangent, the perpendicular from the centre meets the tangent at the point of contact, and this perpendicular is the radius. So radius = 4 cm. The diameter is twice the radius: 2 × 4 = 8 cm. The same idea appears in NCERT Exercise 10.2 Q1, where the radius is found using this right angle.

Key idea to remember

Tangent is perpendicular to the radius through the point of contact: see Key ideas used on this page.

Explanation

This fact of circles says that at the point where a tangent touches a circle, the radius drawn to that point makes a right angle with the tangent. Therefore the shortest distance from the centre to the tangent is exactly the radius, and the diameter is twice it.

Important formulas

Diameter = 2 × radius · Used after finding the radius as the distance from the centre to the tangent.
Radius ⊥ tangent at the point of contact · This is why the distance from the centre to the tangent equals the radius.

Q3. If \(TP\) and \(TQ\) are two tangents to a circle with centre \(O\) from an external point \(T\) so that \(\angle POQ = 120^\circ\) , then \(\angle PTQ\) is equal to :

Mark practised

(A) \(60^\circ\)
(B) \(70^\circ\)
(C) \(80^\circ\)
(D) \(90^\circ\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2026
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2026 Mathematics Standard, Set 30/1/1, Q12, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 60°, because ∠PTQ and ∠POQ are supplementary, so ∠PTQ = 180° − 120° = 60°.
Why? The angle between the two tangents from an external point is supplementary to the angle between the radii at the points of contact, so the two angles add to 180°.

The full 1 mark is for the correct option.

Answer:

(A) 60°

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/1/1: option (A).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 70° because the NCERT exercise uses 110° at the centre; here the centre angle is 120°, so the answer is 60°.
Thinking the angle between the tangents is 120°, the same as the angle at the centre, instead of subtracting it from 180°.
Taking ∠PTQ as 90° just because each tangent is perpendicular to its radius; the angle between the tangents is 60°.

Why this works

Join OP and OQ. The radius at the point of contact is perpendicular to the tangent, so ∠OPT = 90° and ∠OQT = 90°. In quadrilateral OPTQ, the angles add to 360°, so ∠PTQ + 90° + ∠POQ + 90° = 360°, which gives ∠PTQ + ∠POQ = 180°. Put ∠POQ = 120°: ∠PTQ = 180° − 120° = 60°. Quick check: 60° + 120° = 180°, and 90° + 90° + 60° + 120° = 360°.

Key idea to remember

Angle between two tangents drawn from an external point: see Key ideas used on this page.

Explanation

In the chapter 'Circles', the angle between the two tangents drawn from an external point is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. Here P and Q are the points of contact, so ∠PTQ + ∠POQ = 180°. With the centre angle 120°, ∠PTQ = 60°.

Important formulas

∠PTQ + ∠POQ = 180° · The angle between the two tangents from an external point and the angle between the radii at the points of contact are supplementary.

Q4. Four tangents drawn to a circle are extended from both the sides to form a quadrilateral. Which of these quadrilateral is not possible?

Mark practised

(A) Trapezium
(B) Square
(C) Rectangle
(D) Rhombus

Marks: 1
Level: Standard
Years: 2026
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2026 Mathematics Standard, Set 30(B), Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) Rectangle, because in a quadrilateral formed by four tangents the sums of opposite sides are equal, so a rectangle would have length = breadth and therefore must be a square.
Why? The two tangents drawn from an external point to a circle are equal; adding these equal lengths around the quadrilateral gives AB + CD = AD + BC, which a rectangle with unequal sides cannot satisfy.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30(B): option (C).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 'Square' because a square can be drawn around a circle, without checking that 'Rectangle' is the option that fails because it must become a square.
Thinking a rhombus cannot be formed because its angles are not right angles; a parallelogram circumscribing a circle is a rhombus, so it is possible.
Rejecting 'Trapezium' because it is not regular; a trapezium can be formed when the sum of its parallel sides equals the sum of its non-parallel sides.

Why this works

Four tangents form a quadrilateral that circumscribes the circle. From each vertex, the two tangent segments to the circle are equal. Let these equal lengths be a, b, c, d around the quadrilateral. Then the sides are a+b, b+c, c+d and d+a. So the sums of opposite sides are both a+b+c+d, that is, AB + CD = AD + BC. A square and a rhombus satisfy this condition, and a trapezium can also be drawn around a circle. A rectangle would need length = breadth, which is exactly a square. Therefore the quadrilateral that is not possible is the rectangle.

Key idea to remember

Lengths of tangents drawn from an external point to a circle are equal: see Key ideas used on this page.

Explanation

At each vertex of a quadrilateral circumscribing a circle, the two tangent segments to the circle are equal. Label the equal tangent lengths from the four vertices as a, b, c, d. Then the four sides are a+b, b+c, c+d and d+a. Adding opposite sides gives the same total a+b+c+d, so AB + CD = AD + BC. For a rectangle, the opposite sides are equal already; the condition then forces length = breadth. Hence a rectangle must be a square, so a rectangle with unequal sides cannot be formed.

Important formulas

AB + CD = AD + BC · For a quadrilateral circumscribing a circle, the sums of the lengths of opposite sides are equal.

Q5. In two concentric circles with centre O, the radius of outer circle is 25 cm. Chord PQ of the outer circle is tangent to the inner circle at R. If \(PQ = 14\) cm, then the radius of the inner circle is :

Mark practised

(A) \(\sqrt{429}\) cm
(B) 24 cm
(C) \(\sqrt{674}\) cm
(D) 20 cm

Marks: 1
Level: Basic
Years: 2026
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2026 Mathematics Basic, Set 430(B), Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 24 cm, because the perpendicular from O to chord PQ bisects it, so PR = 14/2 = 7 cm, and OR = √(25² − 7²) = √576 = 24 cm.
Why? The radius to the point of contact is perpendicular to the tangent, and the perpendicular from the centre to a chord bisects it, so triangle OPR is right-angled with PR = 7 cm and OP = 25 cm.

The full 1 mark is for the correct option.

Answer:

(B) 24 cm

✓ Answer matches the CBSE answer key, 2026 Main, Paper 430(B): option (B).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Using PR = 14 cm instead of 7 cm; the point of contact R bisects the chord, so each half is 14 ÷ 2 = 7 cm.
Adding the squares and taking the square root gives √674, because 25² + 7² = 674; the correct step is to subtract, since 25 cm is the hypotenuse.
Choosing one of the distractors by guessing, without using the Pythagoras theorem.

Why this works

The circles are concentric, so they have the same centre O. The chord PQ of the outer circle touches the inner circle at R, so PQ is a tangent to the inner circle at R. The radius drawn to the point of contact is perpendicular to the tangent, so OR ⟂ PQ. Now PQ is also a chord of the outer circle, and OR is the perpendicular from the centre O to this chord; such a perpendicular bisects the chord, so PR = RQ = 14 ÷ 2 = 7 cm. In right triangle OPR, the right angle is at R, so the hypotenuse is OP = 25 cm. By the Pythagoras theorem, OR² = OP² − PR² = 25² − 7² = 625 − 49 = 576, so OR = √576 = 24 cm. Check: 24² + 7² = 576 + 49 = 625 = 25².

Key idea to remember

Perpendicular from the centre to a chord bisects the chord: see Key ideas used on this page.

Explanation

Concentric circles have the same centre. For the inner circle, PQ is a tangent at R, so the radius OR is perpendicular to PQ. For the outer circle, PQ is a chord, so the same perpendicular OR from the centre bisects the chord: PR = RQ = 7 cm. Thus the half-chord, the inner radius and the outer radius form a right triangle, and Pythagoras theorem gives the inner radius as 24 cm.

Important formulas

OR² + PR² = OP² · Pythagoras theorem in right triangle OPR; OP is the radius of the outer circle, OR is the radius of the inner circle, and PR is half of PQ.

9 solved questions last asked in 2025

Q6. The distance between two parallel tangents to a circle of radius 3.5 cm is :

Mark practised

(A) 3.5 cm
(B) 14 cm
(C) 1.75 cm
(D) 7 cm

Marks: 1
Level: Basic
Asked in: 3 paper sets
Compartment: 2025
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2025 Compartment Mathematics Basic, Set 430/S/1, Q13, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 7 cm, because the distance between two parallel tangents of a circle is the diameter, 2 × 3.5 = 7 cm.
Why? Each tangent is perpendicular to the radius through its point of contact, so two parallel tangents touch the circle at opposite ends of a diameter, and the distance between them is the diameter.

The full 1 mark is for the correct option.

Answer:

(D) 7 cm

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 430/S/1: option (D).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 3.5 cm because it is the radius, but the question asks for the distance between two parallel tangents, not the radius.
Choosing 14 cm by doubling the diameter; 14 is four times the radius, not twice.
Choosing 1.75 cm by halving the radius, which is half of 3.5 cm and has no connection to the distance between two parallel tangents.

Why this works

Draw a circle and two parallel tangents, one above and one below. The radius drawn to the upper tangent is perpendicular to it, and the radius drawn to the lower tangent is also perpendicular to it. Since the tangents are parallel, these two perpendicular radii lie on the same straight line through the centre, so together they form a diameter. The distance between the two parallel tangents is the length of this perpendicular segment, which is the diameter, 2 × 3.5 = 7 cm. Quick check: 3.5 + 3.5 = 7 cm, so option D is correct.

Key idea to remember

The tangent at any point of a circle is perpendicular to the radius through the point of contact: see Key ideas used on this page.

Explanation

This property says that at the point where a tangent touches a circle, the radius drawn to that point makes a right angle with the tangent. For two parallel tangents, the two perpendicular radii must lie on the same line through the centre, making one full diameter. Hence the distance between the parallel tangents equals the diameter.

Important formulas

Distance between two parallel tangents = 2r · The two tangents touch opposite ends of a diameter, so the distance between them is the diameter.

Q7. A parallelogram having one of its sides 5 cm circumscribes a circle. The perimeter of parallelogram is :

Mark practised

(A) \(20\(\mathrm{cm}\)\)
(B) less than \(20\(\mathrm{cm}\)\)
(C) more than \(20\(\mathrm{cm}\)\) but less than \(40\(\mathrm{cm}\)\)
(D) \(40\(\mathrm{cm}\)\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2025 Mathematics Standard, Set 30/4/1, Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 20 cm, because a parallelogram circumscribing a circle is a rhombus, so each side is 5 cm and perimeter = 4 × 5 = 20 cm.
Why? A parallelogram that circumscribes a circle has all its sides equal, so the perimeter is four times the given side.

The full 1 mark is for the correct option.

Answer:

(A) 20 cm

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/4/1: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Thinking the other two sides of the parallelogram need not be 5 cm, so the perimeter seems to be more than 20 cm.
Forgetting that the tangents drawn from the same external point are equal, and then seeing no way to fix the perimeter.

Why this works

Let ABCD be the parallelogram and let the circle touch AB, BC, CD and DA at P, Q, R and S. From each vertex, the two tangent segments are equal: AP = AS, BP = BQ, CR = CQ and DR = DS. Adding these gives AB + CD = AD + BC. In a parallelogram, AB = CD and AD = BC, so 2AB = 2AD, which gives AB = AD. Thus the parallelogram has all sides equal, so every side is 5 cm. Hence the perimeter is 4 × 5 = 20 cm. Another correct method also earns full marks.

Key idea to remember

Lengths of tangents drawn from an external point to a circle are equal: see Key ideas used on this page.

Explanation

At each vertex of the circumscribing parallelogram, the two sides meeting at that vertex are tangents from that external point, so the tangent lengths are equal. These equalities make the adjacent sides of the parallelogram equal, so the parallelogram becomes a rhombus. Therefore each side is 5 cm and the perimeter is 20 cm.

Important formulas

P = 4a · Perimeter of a rhombus of side a; a parallelogram circumscribing a circle is a rhombus.
AP = AS, BP = BQ, CR = CQ, DR = DS · Equal tangent lengths from each vertex of the circumscribed parallelogram.

Q8. Which of the following statements is false ?

Mark practised

(A) Infinite number of tangents can be drawn to a circle.
(B) Infinite number of tangents can be drawn to a circle from a point outside the circle.
(C) Infinite number of secants can be drawn to a circle from a point outside the circle.
(D) Angle between tangent and diameter at point of contact is 90°.

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 10.2 Tangent to a Circle
Source: (CBSE 2025 Mathematics Standard, Set 30/5/1, Q10, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) Infinite number of tangents can be drawn to a circle from a point outside the circle, because exactly two tangents can be drawn from a point outside a circle.
Why? NCERT states that a point outside a circle has exactly two tangents, so a statement claiming infinitely many tangents from that point contradicts that fact.

The full 1 mark is for the correct option.

Answer:

(B) Infinite number of tangents can be drawn to a circle from a point outside the circle.

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/5/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing A, thinking a circle has only one tangent, although a tangent exists at every point of the circle.
Choosing C, thinking a secant must pass through the centre, although any line cutting the circle in two points is a secant.
Choosing D, thinking the angle is 0° because the diameter passes through the centre, while the tangent is perpendicular to the radius at the point of contact.

Why this works

A tangent touches a circle at exactly one point. A circle has infinitely many points, and at every point there is one tangent, so option A is true. From a point outside a circle, NCERT's section on number of tangents says exactly two tangents can be drawn, so option B is false. A secant cuts a circle at two points; through an outside point, infinitely many such lines can be drawn, so option C is true. By Theorem 10.1, the tangent at a point is perpendicular to the radius through the point of contact, and the diameter through that point lies along the same radius, so the angle between the tangent and the diameter is 90°, making option D true.

Key idea to remember

Number of Tangents from a Point on a Circle: see Key ideas used on this page.

Explanation

NCERT gives three cases: a point inside a circle has no tangent through it, a point on the circle has exactly one tangent, and a point outside the circle has exactly two tangents. Option B says infinitely many tangents can be drawn from an outside point, which contradicts the outside-point case.

Important formulas

Tangent at a point of a circle is perpendicular to the radius through the point of contact · Theorem 10.1; this makes option D true.

Q9. PA and PB are tangents to a circle with centre O. If \(\angle AOB = 105^\circ\) then \(\angle OAP + \angle APB\) is equal to :

Mark practised

(A) \(75^\circ\)
(B) \(175^\circ\)
(C) \(180^\circ\)
(D) \(165^\circ\)

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2025
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2025 Mathematics Basic, Set 430/4/1, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 165°, because ∠OAP = 90° and ∠APB = 360° − 105° − 90° − 90° = 75°, so ∠OAP + ∠APB = 90° + 75° = 165°.
Why? By the tangent theorem, the tangent at a point of a circle is perpendicular to the radius through the point of contact, and the angle sum of a quadrilateral is 360°.

The full 1 mark is for the correct option.

Answer:

(D) 165°

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/4/1: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing only the 75° value of ∠APB, while the question asks for ∠OAP + ∠APB.
Assuming ∠APB has the same value as the given 105° central angle.
Using the angle between the tangent and the radius as 45° instead of a right angle.

Why this works

Join OA and OB. At the point of contact, the tangent and the radius through that point are perpendicular, so ∠OAP = 90° and ∠OBP = 90°. The figure OAPB is a quadrilateral, a four-sided figure, and its angles add up to 360°. With the given ∠AOB of 105°, ∠APB = 360° − 105° − 90° − 90° = 75°. The required sum is ∠OAP + ∠APB = 90° + 75° = 165°. Quick check: for two tangents from the same external point, the angle between the tangents is supplementary to the central angle between the radii to the points of contact, so ∠APB = 180° − 105° = 75°.

Key idea to remember

The tangent at any point of a circle is perpendicular to the radius through the point of contact: see Key ideas used on this page.

Explanation

At each point of contact the tangent is perpendicular to the radius, so two angles of quadrilateral OAPB are right angles. The remaining angle is 360° − 105° − 90° − 90° = 75°, and the required sum is 90° + 75° = 165°.

Important formulas

∠OAP = ∠OBP = 90° · The tangent at a point is perpendicular to the radius through the point of contact.
∠AOB + ∠OAP + ∠OBP + ∠APB = 360° · Angle-sum property of a quadrilateral.
∠APB = 180° − ∠AOB · For two tangents from an external point, the angle between them is supplementary to the central angle between the radii to the points of contact.

Q10. PQ and PR are tangents to the circle of radius 3 cm and centre O. If length of each tangent is 4 cm, then perimeter of \(\Delta\) OQP is :

Mark practised

(A) 5 cm
(B) 12 cm
(C) 9 cm
(D) 8 cm

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2025
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2025 Mathematics Basic, Set 430/4/1, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 12 cm, because OQ = 3 cm, PQ = 4 cm, ∠OQP = 90°, so OP = √(3² + 4²) = 5 cm and perimeter = 3 + 4 + 5 = 12 cm.
Why? The radius through the point of contact is perpendicular to the tangent, so ΔOQP is a right triangle; Pythagoras Theorem gives the missing side.

The full 1 mark is for the correct option.

Answer:

(B) 12 cm

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/4/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 5 cm, because that is only OP, not the perimeter of ΔOQP.
Choosing 8 cm by adding the two tangents 4 + 4; R is not a vertex of ΔOQP and OP is missing.
Taking OP as the radius of 3 cm; OP joins the external point P to the centre O, so it must be found by Pythagoras.

Why this works

P is the external point, and Q and R are the points of contact. The circle has radius 3 cm, so OQ = OR = 3 cm. Each tangent has length 4 cm, so PQ = PR = 4 cm. At the point of contact Q, the radius OQ is perpendicular to the tangent PQ, so ΔOQP is right-angled at Q. By Pythagoras Theorem, OP² = 3² + 4² = 25, so OP = 5 cm. Hence perimeter of ΔOQP = 3 + 4 + 5 = 12 cm. A quick check: 3, 4, 5 is a Pythagorean triple, so OP = 5 cm is consistent. The wrong options are partial lengths: 5 cm is only OP, and 8 cm is PQ + PR.

Key idea to remember

The tangent at any point of a circle is perpendicular to the radius through the point of contact: see Key ideas used on this page.

Explanation

The tangent at any point of a circle is perpendicular to the radius through the point of contact. At Q, the radius OQ and the tangent QP meet at a right angle, so ΔOQP is a right triangle. Pythagoras Theorem gives OP = 5 cm, and then the perimeter of ΔOQP is 3 + 4 + 5 = 12 cm.

Important formulas

OP² = OQ² + QP² · Pythagoras Theorem applied to right ΔOQP.
Perimeter of a triangle = sum of its three sides · Perimeter of ΔOQP = OQ + QP + OP.

Q11. Rectangle ABCD circumscribes the circle of radius 10 cm. Prove that ABCD is a square. Hence, find the perimeter of ABCD.

Mark practised

Marks: 3
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2025 Mathematics Standard, Set 30/4/1, Q31, 3 marks)
Status: Verified solution

Step-by-step solution

Step 1: Mark the points of contact and equal tangents
1 mark (suggested)
Let the circle touch AB, BC, CD, DA at P, Q, R, S respectively. The lengths of tangents drawn from an external point to a circle are equal, so AP = AS, BP = BQ, CQ = CR, DR = DS.
Why? From each corner of the rectangle, two tangent segments go to the circle, and the theorem says these two lengths are equal.
Step 2: Compare opposite sides and conclude square
1 mark (suggested)
Therefore AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS) = (AS + DS) + (BQ + CQ) = AD + BC. Since ABCD is a rectangle, AB = CD and AD = BC, so 2AB = 2AD, hence AB = AD. Therefore ABCD is a square.
Why? For a quadrilateral circumscribing a circle, the sums of opposite sides are equal. In a rectangle, opposite sides are equal, so this forces adjacent sides to be equal, which is exactly what makes a rectangle a square.
Step 3: Find the side and the perimeter
1 mark (suggested)
The distance between the parallel tangents AB and CD is the side of the square, and it equals the diameter 2 × 10 = 20 cm. Hence the perimeter of ABCD is 4 × 20 = 80 cm.
Why? The radius to a tangent is perpendicular to it, so the distance between two parallel tangents is the diameter. The perimeter of a square is four times its side.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

Answer:

80 cm

✓ Answer matches the CBSE marking scheme, 2025 Main, Paper 30/4/1.

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Using the radius 10 cm as the side of the square, which gives 40 cm instead of 80 cm; the side is the diameter 20 cm.
Stopping after writing AB + CD = AD + BC, without using AB = CD and AD = BC to get AB = AD.
Forgetting that a rectangle with equal adjacent sides is a square, and so not stating the conclusion clearly.

Why this works

Circumscribing the circle means each side of the rectangle is a tangent to the circle. Name the four points of contact. From each corner of the rectangle, two tangents go to the circle, so the two tangent lengths from that corner are equal. Adding the four equalities shows that in a quadrilateral circumscribing a circle the sums of opposite sides are equal. In a rectangle, opposite sides are equal, so this gives AB = AD; a rectangle with equal adjacent sides is a square. The side of this square is the distance between the two parallel tangents AB and CD, which is the diameter 2 × 10 = 20 cm. Hence the perimeter is 4 × 20 = 80 cm. Quick check: 2 × 10 = 20 and 4 × 20 = 80.

Key idea to remember

Lengths of tangents drawn from an external point to a circle are equal: see Key ideas used on this page.

Explanation

For any outside point, the two tangent segments to a circle have the same length. Here the points A, B, C, D are outside the circle, so AP = AS, BP = BQ, CQ = CR and DR = DS. Adding these equalities lets us compare opposite sides of the rectangle.

Important formulas

AP = AS, BP = BQ, CQ = CR, DR = DS · Equal tangents drawn from an external point.
AB + CD = AD + BC · Sums of opposite sides of a quadrilateral circumscribing a circle.
d = 2r · The side of the square is the diameter of the circle; the radius is 10 cm.
P = 4s · Perimeter of a square of side s.

Q12. The tangents drawn at the extremities of the diameter of a circle are always :

Mark practised

(A) parallel
(B) perpendicular
(C) equal
(D) intersecting

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2025 Mathematics Standard, Set 30/1/1, Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) parallel, because each tangent is perpendicular to the radius at its point of contact and the two radii lie on the same diameter.
Why? By Theorem 10.1, the tangent at any point of a circle is perpendicular to the radius through the point of contact. At the two ends of a diameter, both radii lie on the same straight line, so the two tangents are perpendicular to the same line and hence are parallel.

The full 1 mark is for the correct option.

Answer:

(A) parallel

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/1/1: option (A).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 'perpendicular' because each tangent is at right angles to a radius, without checking the relation between the two tangents.
Choosing 'intersecting' without noticing that both tangents are perpendicular to the same diameter, so they never meet.
Choosing 'equal' because tangent lengths from an external point are equal, which is a different situation from tangents at the ends of a diameter.

Why this works

Let AB be a diameter of the circle with centre O. The tangent at A is perpendicular to OA, and the tangent at B is perpendicular to OB. Since OA and OB are parts of the same diameter AB, both tangents are perpendicular to the same line AB. Two lines perpendicular to the same line are parallel, so the tangents are parallel. This is exactly the statement proved in NCERT Exercise 10.2 Q4. The other options are wrong: the tangents do not meet, so they cannot be intersecting or perpendicular to each other; and 'equal' refers to lengths of tangents from an external point, which is not the situation here.

Key idea to remember

Tangent perpendicular to radius through the point of contact: see Key ideas used on this page.

Explanation

Theorem 10.1 states that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Applying this at both ends of a diameter gives two tangents perpendicular to the same diameter, so they are parallel to each other.

Important formulas

Tangent at a point is perpendicular to the radius through the point of contact. · Used at both ends of the diameter.
Two lines perpendicular to the same line are parallel. · This gives the final conclusion.

Q13. For a circle with centre O and radius 5 cm, which of the following statements is true? 1
P : Distance between every pair of parallel tangents is 5 cm.
Q : Distance between every pair of parallel tangents is 10 cm.
R : Distance between every pair of parallel tangents must be between 5 cm and 10 cm.
S : There does not exist a point outside the circle from where length of tangent is 5 cm.

Mark practised

(A) P
(B) Q
(C) R
(D) S

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2025 Mathematics Standard, Set 30/6/1, Q10, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) Q, because the distance between two parallel tangents is the diameter, 2 × 5 = 10 cm.
Why? Each tangent is perpendicular to the radius at its point of contact; two parallel tangents touch the circle at the two ends of a diameter, so the distance between them is the diameter.

The full 1 mark is for the correct option.

Answer:

(B) Q

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/6/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing P because 5 cm is the radius, without remembering that parallel tangents are separated by the full diameter, 10 cm.
Choosing R, thinking the distance between parallel tangents can be any value between 5 cm and 10 cm, when it is fixed at 10 cm.
Choosing S, thinking no external point can give a tangent of length 5 cm, although a point at distance 5√2 cm from the centre does give a 5 cm tangent.

Why this works

A tangent at any point is perpendicular to the radius through that point. Two parallel tangents are perpendicular to the same direction, so the radii to their points of contact lie on one straight line through O. Hence the points of contact are the two ends of a diameter, and the distance between the tangents is the diameter: 2 × 5 = 10 cm. So Q is true. P is false because 5 cm is the radius, not the distance between parallel tangents. R is false because this distance is fixed at 10 cm, not a range from 5 cm to 10 cm. S is false: choose an external point P with OP² = 50 cm²; then the tangent length is √(OP² − 5²) = √(50 − 25) = 5 cm, so such a point exists.

Key idea to remember

The tangent at any point of a circle is perpendicular to the radius through the point of contact.: see Key ideas used on this page.

Explanation

This property fixes where a tangent touches the circle. Two parallel tangents are perpendicular to the same line, so the two radii through their points of contact form one diameter. Therefore the perpendicular distance between the tangents equals the diameter, which is twice the radius.

Important formulas

distance between two parallel tangents = 2r · For this circle, r = 5 cm, so the distance is 10 cm.
PT² = OP² − r² · Length PT of a tangent from an external point P to a circle of radius r, using Pythagoras.

Q14. For a circle with centre O and radius 5 cm, which of the following statements is true ?
P : Distance between every pair of parallel tangents is 5 cm.
Q : Distance between every pair of parallel tangents is 10 cm.
R : Distance between every pair of parallel tangents must be between 5 cm and 10 cm.
S : There does not exist a point outside the circle from where length of tangent is 5 cm.

Mark practised

(A) P
(B) Q
(C) R
(D) S

Marks: 1
Level: Standard
Years: 2025
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2025 Mathematics Standard, Set 30/6/2, Q13, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) Q, because parallel tangents touch the circle at opposite ends of a diameter, so their distance is 2 × 5 = 10 cm.
Why? The radius through the point of contact is perpendicular to the tangent, so the two radii for parallel tangents lie on one straight line and the distance between the tangents is the diameter, 2r.

The full 1 mark is for the correct option.

Answer:

(B) Q

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/6/2: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing P because the radius is 5 cm and forgetting that the distance between parallel tangents is the full diameter, 10 cm.
Thinking R is correct because the distance between parallel tangents changes with direction; it is always the diameter, 10 cm.
Thinking S is true because the radius is 5 cm, without checking that an outside point at distance √50 cm gives a tangent of length 5 cm.

Why this works

For any given direction, a circle has exactly two parallel tangents, one on each side. At the point of contact, the radius is perpendicular to the tangent. So the two radii at the two points of contact lie on the same straight line through the centre. That line is a diameter, and the distance between the parallel tangents equals this diameter, 2 × 5 = 10 cm. Hence Q is true, while P and R are false. For S, choose an external point T with OT² = 5² + 5² = 50, so OT is √50 cm. If P is the point of contact of a tangent from T, then OP is perpendicular to PT, so PT² = OT² − OP² = 50 − 25 = 25, giving PT = 5 cm. Thus a point outside the circle with tangent length 5 cm does exist, so S is false. Only Q is true.

Key idea to remember

Distance between parallel tangents to a circle: see Key ideas used on this page.

Explanation

For a given direction, the two parallel tangents touch the circle at the ends of a diameter. Because the radius at the point of contact is perpendicular to the tangent, the two radii lie on one straight line, and the distance between the tangents is the diameter, 2r.

Important formulas

Distance between parallel tangents = 2r · For this circle the radius is 5 cm, so the distance is 10 cm.
PT² = OT² − r² · Length PT of a tangent from an external point T to a circle of radius r with centre O; used to check statement S.

3 solved questions last asked in 2024

Q15. A line intersecting a circle in two distinct points is called a

Mark practised

(A) secant
(B) chord
(C) diameter
(D) tangent

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2024
Topic: 10.2 Tangent to a Circle
Source: (CBSE 2024 Mathematics Basic, Set 430/2/2, Q12, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) secant, because this line meets the circle at 2 distinct points, which is the defining property of a secant.
Why? A secant of a circle is defined as a line that has 2 common points with the circle.

The full 1 mark is for the correct option.

Answer:

(A) secant

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/2/2: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 'chord' because a chord also joins two points of a circle, but a chord is the segment between the two points, while the question asks for the whole line.
Choosing 'tangent' because the line touches the circle, but a tangent has exactly one common point with the circle, not two.
Choosing 'diameter' because a diameter passes through two points, but a diameter is a chord through the centre, not just any line intersecting the circle at two points.

Q16. A tangent to a circle is a line that touches the circle at :

Mark practised

(A) one point only
(B) two points
(C) three points
(D) infinite number of points

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2024
Topic: 10.2 Tangent to a Circle
Source: (CBSE 2024 Mathematics Basic, Set 430/4/1, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) one point only, because a tangent to a circle intersects the circle at only one point.
Why? A tangent is defined as the line that has exactly one common point with the circle; that common point is called the point of contact.

The full 1 mark is for the correct option.

Answer:

(A) one point only

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/4/1: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing B, two points, because that describes a secant, not a tangent.
Thinking a tangent touches the circle at more than one point; a line cannot touch a circle at two points and still be a tangent.

Q17. From a point P which is at a distance of 17 cm from the centre O of a circle of radius 8 cm, the pair of tangents PQ and PR to the circle are drawn. The area of triangle OPQ is :

Mark practised

(A) 120 sq cm
(B) 68 sq cm
(C) 60 sq cm
(D) 127.5 sq cm

Marks: 1
Level: Basic
Compartment: 2024
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2024 Compartment Mathematics Basic, Set 430(B), Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 60 sq cm, because PQ = √(17² − 8²) = 15 cm and area = 1/2 × 15 × 8 = 60 sq cm.
Why? The tangent at any point of a circle is perpendicular to the radius through the point of contact, so ΔOPQ is right-angled at Q. Pythagoras theorem gives PQ, and the area of a right triangle is half the product of its legs.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Compartment, Paper 430(B): option (C).

Open CBSE’s official marking scheme for the 2024 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 68 sq cm by taking half of 17 × 8, which treats OP and OQ as the two perpendicular legs; OP is the hypotenuse, not a leg.
Choosing 120 sq cm by computing 15 × 8 without halving.
Choosing 127.5 sq cm by taking half of 17 × 15, treating OP as perpendicular to PQ; only OQ is perpendicular to PQ.

Why this works

Q is the point of contact, so OQ is a radius and OQ ⊥ PQ. Hence ΔOPQ is right-angled at Q, with hypotenuse OP. The given OP is 17 cm and the radius OQ is 8 cm. By Pythagoras theorem, PQ² = OP² − OQ² = 17² − 8² = 225, so PQ = 15 cm. Area of ΔOPQ = 1/2 × PQ × OQ = 1/2 × 15 × 8 = 60 sq cm. Therefore option C is correct. Quick check: 17² = 289, 8² = 64 and 289 − 64 = 225 = 15².

Key idea to remember

The tangent at any point of a circle is perpendicular to the radius through the point of contact: see Key ideas used on this page.

Explanation

When a tangent touches a circle, the radius drawn to the point of contact makes a right angle with the tangent. This turns ΔOPQ into a right triangle, so Pythagoras theorem and the triangle area formula can be applied.

Important formulas

PQ² = OP² − OQ² · Pythagoras theorem in right triangle OPQ, where OP is the hypotenuse and OQ is the radius.
Area = 1/2 × base × height · For triangle OPQ, take base PQ and height OQ.

5 solved questions last asked in 2023

Q18. The length of the tangent drawn from a point P, whose distance from the centre of a circle is 25 cm, and the radius of the circle is 7 cm, is :

Mark practised

(A) 22 cm
(B) 24 cm
(C) 25 cm
(D) 28 cm

Marks: 1
Level: Standard
Asked in: 3 paper sets
Compartment: 2023
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2023 Compartment Mathematics Standard, Set 30/C/1, Q15, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 24 cm, because the tangent is perpendicular to the radius, so by Pythagoras theorem √(25² − 7²) = 24 cm.
Why? The tangent at the point of contact is perpendicular to the radius, so the radius, the tangent segment and the line joining the external point to the centre form a right triangle with the distance from the centre as the hypotenuse.

The full 1 mark is for the correct option.

Answer:

(B) 24 cm

✓ Answer matches the CBSE answer key, 2023 Compartment, Paper 30/C/1: option (B).

Open CBSE’s official marking scheme for the 2023 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 32 cm by adding 25 and 7, forgetting that OP is the hypotenuse, not a leg.
Choosing 18 cm by subtracting 7 from 25, without squaring the lengths.
Finding 25² − 7² = 576 and forgetting to take the square root, so giving 576 cm instead of 24 cm.

Why this works

Draw the circle with centre O and point P outside it. Let the tangent touch the circle at T. Then OT is a radius of 7 cm, OP = 25 cm, and ∠OTP = 90° because the tangent is perpendicular to the radius. So triangle OTP is right-angled at T, and OP is the hypotenuse. By Pythagoras theorem, PT² = OP² − OT² = 25² − 7² = 625 − 49 = 576, so PT = 24 cm. This is the same right-triangle setup as NCERT Exercise 10.2 Q1, where the tangent length and the distance from the centre are given and the radius is asked.

Key idea to remember

The tangent at any point of a circle is perpendicular to the radius through the point of contact: see Key ideas used on this page.

Explanation

At the point where a tangent touches a circle, the radius drawn to that point makes a right angle with the tangent. Therefore the radius, the tangent segment and the line from the external point to the centre form a right triangle, and the unknown length can be found by the Pythagoras theorem.

Important formulas

PT² = OP² − OT² · Pythagoras theorem applied to the right triangle formed by the radius, the tangent and the line from the external point to the centre.

Q19. The distance between two parallel tangents of a circle of diameter 7 cm is :

Mark practised

(A) 7 cm
(B) 14 cm
(C) \(\frac{7}{2}\) cm
(D) 28 cm

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2023
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2023 Mathematics Basic, Set 430/1/2, Q13, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 7 cm, because the distance between two parallel tangents of a circle is equal to its diameter, 7 cm.
Why? The tangent at any point of a circle is perpendicular to the radius through the point of contact, so the points of contact of two parallel tangents are the ends of a diameter.

The full 1 mark is for the correct option.

Answer:

(A) 7 cm

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/1/2: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 14 cm by treating 7 cm as the radius, but 7 cm is already the diameter.
Choosing 7/2 cm, because that is the radius, not the distance between the two parallel tangents.
Choosing 28 cm by doubling the diameter again.

Why this works

Draw a circle with two parallel tangents, one above and one below. Each tangent touches the circle at exactly one point. The radius drawn to the upper point of contact is perpendicular to the upper tangent, so it is vertical; the radius to the lower point is also vertical. Both radii pass through the centre, so the two points of contact are the two ends of one diameter. The distance between the two parallel lines is the distance between these two points, which is the diameter. Since the diameter is 7 cm, the distance between the parallel tangents is 7 cm. The radius is 7/2 cm, which is only the distance from the centre to either tangent.

Key idea to remember

The tangent at any point of a circle is perpendicular to the radius through the point of contact: see Key ideas used on this page.

Explanation

This property forces the two points of contact of parallel tangents to lie on the same diameter. Therefore the distance between the parallel tangents equals the diameter of the circle.

Important formulas

distance between two parallel tangents = 2r · The two points of contact are the ends of a diameter, so the distance between the tangents is the diameter.

Q20. How many tangents can be drawn to a circle from a point on it ?

Mark practised

(A) One
(B) Two
(C) Infinite
(D) Zero

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2023
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2023 Mathematics Basic, Set 430/4/1, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) One, because exactly one tangent can be drawn through a point on the circle.
Why? A point lying on a circle has one and only one tangent, and this tangent is perpendicular to the radius through the point of contact.

The full 1 mark is for the correct option.

Answer:

(A) One

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/4/1: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing Two because two tangents can be drawn from a point outside the circle, but this point is on the circle.
Choosing Infinite because a circle has infinitely many tangents in total, but only one passes through a fixed point on the circle.
Choosing Zero because a point inside the circle has no tangent, but this point is on the circle.

Q21. A circle is of radius 3 cm. The distance between two of its parallel tangents is :

Mark practised

(A) 12 cm
(B) 6 cm
(C) 3 cm
(D) 4.5 cm

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2023
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2023 Mathematics Basic, Set 430/5/1, Q14, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 6 cm, because the distance between two parallel tangents equals the diameter, 2 × 3 = 6 cm.
Why? Parallel tangents touch the circle at opposite ends of a diameter, so the distance between them is the diameter.

The full 1 mark is for the correct option.

Answer:

(B) 6 cm

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/5/1: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 3 cm because it is the radius, but the gap between two parallel tangents is the full diameter.
Choosing 12 cm by taking four times the radius, but the correct distance is two times the radius.
Thinking parallel tangents can lie on the same side of the circle; two parallel tangents must touch opposite ends of a diameter.

Why this works

A tangent touches a circle at exactly one point. At that point the radius is perpendicular to the tangent. A circle has only two tangents parallel to a given line: one on each side. They touch the circle at the two ends of a diameter. The distance between two parallel lines is the length of a perpendicular segment between them, so here it is the diameter. The given radius is 3 cm, so the distance is 2 × 3 = 6 cm. The other options are not possible: 12 cm is four radii, 3 cm is only the radius, and 4.5 cm is not the diameter.

Key idea to remember

Tangent to a Circle: see Key ideas used on this page.

Explanation

In the NCERT chapter, a tangent is a line that touches the circle at only one point, and the radius through the point of contact is perpendicular to the tangent. For two parallel tangents, the two radii drawn to their points of contact lie on the same straight line, making the distance between the tangents equal to the diameter of the circle.

Important formulas

distance between parallel tangents = 2r · where r is the radius of the circle
diameter = 2r · used with the given radius of 3 cm

Q22. The length of the tangent from an external point A to a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is :

Mark practised

(A) 7 cm
(B) 5 cm
(C) √7 cm
(D) 25 cm

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2023
Topic: 10.3 Number of Tangents from a Point on a Circle
Source: (CBSE 2023 Mathematics Basic, Set 430/4/2, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 5 cm, because the tangent is perpendicular to the radius, so OA = √(3² + 4²) = 5 cm.
Why? The radius drawn to the point of contact is perpendicular to the tangent, so the radius, the tangent and the distance OA form a right triangle.

The full 1 mark is for the correct option.

Answer:

(B) 5 cm

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/4/2: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 7 cm by adding 3 cm and 4 cm, instead of using Pythagoras theorem.
Choosing √7 cm by writing 4² − 3², but OA is the hypotenuse, not a leg.
Choosing 25 cm because 3² + 4² gives 25, without taking the square root.

Why this works

Let O be the centre of the circle and let the tangent from A touch the circle at P. Then OP is a radius of length 3 cm, and AP is the tangent of length 4 cm. The radius OP is perpendicular to the tangent AP, so triangle OPA is right-angled at P. By Pythagoras theorem, OA² = 3² + 4² = 25, so OA = 5 cm. The distance of A from the centre is 5 cm.

Key idea to remember

Tangent to a circle is perpendicular to the radius through the point of contact: see Key ideas used on this page.

Explanation

When a tangent touches a circle at a point, the radius drawn to that point makes a right angle with the tangent. This creates a right triangle whose hypotenuse is the distance of the external point from the centre, so Pythagoras theorem gives the required distance.

Important formulas

OA² = OP² + AP² · In right triangle OPA, OA is the distance of the external point A from the centre O, OP is the radius, and AP is the tangent length.

Most repeated Circles questions

The questions CBSE printed in the most question-paper sets, 2023–2026. Each one is solved where the line points.

Printed in 6 question-paper sets (2025 and 2026) · 2 marks · Basic · question 1 on this page
Printed in 3 question-paper sets (2023 compartment) · 1 mark · Standard · question 18 on this page
Printed in 3 question-paper sets (2023) · 1 mark · Basic · question 19 on this page
Printed in 3 question-paper sets (2023) · 1 mark · Basic · question 20 on this page
Printed in 3 question-paper sets (2023) · 1 mark · Basic · question 21 on this page

Official papers behind this chapter

Circles was asked in 120 of the 120 question-paper sets of 2023–2026. The table names the official CBSE files for each exam in which it was asked.

The board publishes the question papers on cbse.gov.in and the marking schemes on cbse.gov.in, where you can see how marks are given.

Year	Exam	Course	Sets with a Circles question	Question paper	Marking scheme
2026	Main	Standard (041)	10 of 10 question-paper sets had a Circles question	Open CBSE's official question paper for the 2026 main exam (one ZIP file of about 61.9 MB, Class 10 Maths Standard and Basic together) to read the full paper.	Open CBSE's official marking scheme for the 2026 main exam (one ZIP file of about 22.8 MB, Class 10 Maths Standard (041)) to see how marks are given.
2026	Main	Basic (241)	10 of 10 question-paper sets had a Circles question	Open CBSE's official question paper for the 2026 main exam (one ZIP file of about 61.9 MB, Class 10 Maths Standard and Basic together) to read the full paper.	Open CBSE's official marking scheme for the 2026 main exam (one ZIP file of about 29.1 MB, Class 10 Maths Basic (241)) to see how marks are given.
2025	Main	Standard (041)	17 of 17 question-paper sets had a Circles question	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 54.5 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Main	Basic (241)	12 of 12 question-paper sets had a Circles question	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 49.9 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Compartment	Standard (041)	3 of 3 question-paper sets had a Circles question	Open CBSE's official question paper for the 2025 compartment exam (one ZIP file of about 3.4 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2025 compartment exam (one ZIP file of about 8.1 MB, Class 10 Maths Standard (041)) to see how marks are given.
2025	Compartment	Basic (241)	3 of 3 question-paper sets had a Circles question	Open CBSE's official question paper for the 2025 compartment exam (one ZIP file of about 2.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 compartment exam (one ZIP file of about 18.7 MB, Class 10 Maths Basic (241)) to see how marks are given.
2024	Main	Standard (041)	14 of 14 question-paper sets had a Circles question	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 35.3 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2024	Main	Basic (241)	14 of 14 question-paper sets had a Circles question	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 43.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.8 MB, Class 10 Maths Basic (241)) to see how marks are given.
2024	Compartment	Standard (041)	3 of 3 question-paper sets had a Circles question	Open CBSE's official question paper for the 2024 compartment exam (one ZIP file of about 3.2 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2024 compartment exam (one ZIP file of about 7.1 MB, Class 10 Maths Standard (041)) to see how marks are given.
2024	Compartment	Basic (241)	4 of 4 question-paper sets had a Circles question	Open CBSE's official question paper for the 2024 compartment exam (one ZIP file of about 2.2 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 compartment exam (one ZIP file of about 5.6 MB, Class 10 Maths Basic (241)) to see how marks are given.
2023	Main	Standard (041)	12 of 12 question-paper sets had a Circles question	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 44 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 17.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2023	Main	Basic (241)	12 of 12 question-paper sets had a Circles question	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 49.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 26 MB, Class 10 Maths Basic (241)) to see how marks are given.
2023	Compartment	Standard (041)	3 of 3 question-paper sets had a Circles question	Open CBSE's official question paper for the 2023 compartment exam (one ZIP file of about 7.6 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2023 compartment exam (one ZIP file of about 3.3 MB, Class 10 Maths Standard (041)) to see how marks are given.
2023	Compartment	Basic (241)	3 of 3 question-paper sets had a Circles question	Open CBSE's official question paper for the 2023 compartment exam (one ZIP file of about 7.2 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2023 compartment exam (one ZIP file of about 7.5 MB, Class 10 Maths Basic (241)) to see how marks are given.

Continue studying

Other chapters: Areas Related to Circles previous year questions · Some Applications of Trigonometry previous year questions.

To revise the lessons behind these questions, read NCERT Chapter 10 (official PDF), the chapter exactly as NCERT publishes it on ncert.nic.in.

Sources and data verification

The counts on this page come from 120 CBSE Class 10 Mathematics question-paper sets: 62 Standard (041) and 58 Basic (241), from the main exams of 2023, 2024, 2025 and 2026 and the compartment exams of 2023, 2024 and 2025. A question printed in several sets counts once; these counts describe past papers and are not predictions.

CBSE settles the question papers and the marking schemes: each question is reproduced as printed in its CBSE paper, with its year, paper code and question number, and each answer shown here matches the official marking scheme or answer key. NCERT settles the textbook: chapter and section names follow NCERT Mathematics Class X (reprint 2026-27).

Every solution is original. The mark beside each step follows the official CBSE marking scheme where the scheme shows it; a mark labelled “(suggested)” follows the board’s usual pattern for that type of question.

The official question-paper and marking-scheme links on this page were checked and open on 9 October 2026.

Reference: CBSE Class 10 Mathematics board question papers via cbse.gov.in; all solutions are original. learncbse.net is an independent website and is not affiliated with CBSE or NCERT.

Circles in CBSE board papers

Where Circles questions came from, 2023–2026

Questions by exam

Questions by NCERT section and marks

Most-tested sections

Topics in Circles

10.2 Tangent to a Circle

How to answer questions on this section

Mistakes to avoid in this section

10.3 Number of Tangents from a Point on a Circle

How to answer questions on this section

Mistakes to avoid in this section

Key ideas used on this page

5 solved questions last asked in 2026

Q1. Two concentric circles are of radii 6 cm and 10 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Step-by-step solution

Step 1: Right angle at point of contact

Step 2: Find half the chord

Step 3: Length of the chord

Answer:

Common mistakes to avoid

Q2. If the distance of a tangent to a circle from its centre is 4 cm, then the length of diameter of the circle is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q3. If \(TP\) and \(TQ\) are two tangents to a circle with centre \(O\) from an external point \(T\) so that \(\angle POQ = 120^\circ\) , then \(\angle PTQ\) is equal to :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q4. Four tangents drawn to a circle are extended from both the sides to form a quadrilateral. Which of these quadrilateral is not possible?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q5. In two concentric circles with centre O, the radius of outer circle is 25 cm. Chord PQ of the outer circle is tangent to the inner circle at R. If \(PQ = 14\) cm, then the radius of the inner circle is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

9 solved questions last asked in 2025

Q6. The distance between two parallel tangents to a circle of radius 3.5 cm is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q7. A parallelogram having one of its sides 5 cm circumscribes a circle. The perimeter of parallelogram is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q8. Which of the following statements is false ?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q9. PA and PB are tangents to a circle with centre O. If \(\angle AOB = 105^\circ\) then \(\angle OAP + \angle APB\) is equal to :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q10. PQ and PR are tangents to the circle of radius 3 cm and centre O. If length of each tangent is 4 cm, then perimeter of \(\Delta\) OQP is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q11. Rectangle ABCD circumscribes the circle of radius 10 cm. Prove that ABCD is a square. Hence, find the perimeter of ABCD.

Step-by-step solution

Step 1: Mark the points of contact and equal tangents

Step 2: Compare opposite sides and conclude square

Step 3: Find the side and the perimeter

Answer:

Common mistakes to avoid

Q12. The tangents drawn at the extremities of the diameter of a circle are always :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Step-by-step solution