Looking for the right NCERT solution for Class 10 Maths Chapter 10? This page maps the whole chapter: it tells you which exercise your Circles homework question sits in and which method that exercise expects you to use. The full worked solutions live one level down, on the page for each exercise.
The chapter runs on one fact: a tangent touches a circle at exactly one point. A line can sit against a circle in three ways — with no common point, with two common points (a secant), or with one common point (a tangent). Everything else hangs on two theorems, stated below.
What this chapter is for
Circles (pages 145–153) is a short chapter with a narrow job: it sets up the vocabulary — tangent, secant, point of contact, length of a tangent — and proves the two theorems that make every question in it solvable. Fig. 10.6 shows the three cases you must know cold:
- a point inside the circle gives no tangent
- a point on the circle gives exactly one tangent
- a point outside the circle gives exactly two tangents
Theorem 10.1 (p.147): the tangent at any point of a circle is perpendicular to the radius through the point of contact. Theorem 10.2 (p.149): the lengths of tangents drawn from an external point to a circle are equal. Once you know where your question sits, open that exercise’s page for the working.
Which exercise has your question
Two exercises cover the chapter, and they split cleanly: Exercise 10.1 handles vocabulary and one quick length; Exercise 10.2 handles angles and proofs.
Exercise 10.1 (page 152, four questions) tests the definition of a tangent and the related terms — secant, point of contact. Its one length question expects you to draw the radius that meets the tangent, form the Pythagorean right triangle from Theorem 10.1, and apply \( \text{(hypotenuse)}^2 = \text{(radius)}^2 + \text{(tangent)}^2 \).
Exercise 10.2 (pages 152–153, thirteen questions) opens with three multiple-choice questions on tangent length and tangent-pair angles, then runs into a long proof set: tangents at the ends of a diameter, the perpendicular through the point of contact, concentric circles, and circumscribed polygons.
The same idea — two tangents from an external point are equal — is recycled through to the quadrilateral result \( AB + CD = AD + BC \).
The two theorems that do all the work
Three techniques cover everything the chapter asks, and each exercise drills its own:
- Radius perpendicular to tangent (Theorem 10.1). Every length question — the one in Exercise 10.1 and the tangent-length questions of Exercise 10.2 — becomes a right-triangle problem with the radius and tangent as legs. This is also the angle you lean on inside Exercise 10.2’s proofs.
- Equal tangent lengths (Theorem 10.2). The workhorse of Exercise 10.2. The Pythagoras route \( PQ^2 = OP^2 – OQ^2 \) proves it, and the same congruence shows the centre lies on the bisector of the angle between the two tangents.
- Perpendicular from the centre bisects a chord. Example 1 adds this third tool: when a chord of a larger circle touches a smaller concentric circle, the point of contact is the midpoint of the chord.
Where students trip on tangents
Three errors cost real marks in this chapter:
- The right angle holds only at the point of contact. A perpendicular dropped from the centre to any line is not automatically a radius–tangent relation. The 90° of Theorem 10.1 sits at the exact point where the tangent touches the circle — locate that point before claiming the angle.
- Equal tangent lengths apply only from an external point. Theorem 10.2 needs the point outside the circle. From a point inside, no tangent exists at all, so there is no tangent length to compare.
- The tangent-pair angle is supplementary to the central angle. When the centre and the two points of contact give \( \angle POQ = 110^\circ \), the angle between the tangents is \( 70^\circ \) — the two add to \( 180^\circ \). Learn this relation rather than deriving it under pressure.
How to use this page for homework
Run a three-point check when you open a question:
- A definition, a fill-in-the-blank, or a straight length with the tangent already drawn → Exercise 10.1.
- A proof of anything — parallel tangents, a perpendicular through the centre, a rhombus, \( AB + CD = AD + BC \) — → Exercise 10.2.
- A question that names a figure (Fig. 10.11, 10.12, 10.13 or 10.14) → open the linked exercise page, which shows that figure before the working.
For chapter-level revision, the class 10 maths notes keep these theorems alongside the other chapters, while the previous chapter applies trigonometry and chapter 11 turns circles into area problems. To browse any class or subject, start from the full notes index or the class 10 hub.
For the official wording of both theorems, the NCERT Class 10 Mathematics textbook on ncert.nic.in carries the whole chapter.
| Exercise | What you get |
|---|---|
| Exercise 10.1 | Solved |
| Exercise 10.2 | Solved |
Circles chapter FAQs
How many tangents can a circle have?
Infinitely many — exactly one at every point on the circumference. From a single point outside the circle, exactly two can be drawn.
What is the difference between a tangent and a secant?
A tangent meets the circle at exactly one point, the point of contact. A secant cuts the circle at two points. A tangent is the limiting position of a secant as its two intersection points come together.
In which exercise do I prove that tangents at the ends of a diameter are parallel?
That is a proof question, so it sits in Exercise 10.2 rather than 10.1. The tool is Theorem 10.1 — each tangent is perpendicular to the radius at its point of contact.
How do I know whether a Circles question wants Pythagoras or angle chasing?
If the question hands you two lengths of the tangent triangle and asks for the third, use Pythagoras on the right triangle from Theorem 10.1. If it gives or asks for angles, chase angles with Theorem 10.2 and the supplementary relation. A question that only states relationships is a proof.
Reference: NCERT textbooks (CBSE).