This sheet covers the formulas and results from the NCERT Class 9 chapter “I’m Up and Down, and Round and Round” (Circles): definitions of a circle, chord and arc; the chord–angle and chord–distance theorems; the angle-subtended-by-an-arc results; and the cyclic quadrilateral property.
Each formula below is written in MathJax with every symbol and its unit explained, so you can confirm what each letter means and which situation it applies to.
Every formula is grouped by the textbook sub-topic it belongs to, followed by a symbol table, when-to-use guidance, and three original worked examples. For the full explanations and derivations of each theorem, see the Class 9 maths notes for this chapter.
Formulas at a Glance
| Purpose (what you are finding) | Formula |
|---|---|
| Chord length from radius and perpendicular distance to centre | \( \text{Chord length} = 2\sqrt{r^2 – d^2} \) |
| Distance of a chord from the centre (from the chord-length formula, when \( r \gt d \)) | \( d = \sqrt{r^2 – \left(\frac{\text{chord length}}{2}\right)^2} \) |
| Radius from chord length and distance from centre | \( r = \sqrt{d^2 + \left(\frac{\text{chord length}}{2}\right)^2} \) |
| Angle subtended by an arc at the centre is double the angle at any point on the remaining circle | \( \angle \text{at centre} = 2 \times \angle \text{at a point on the circle} \) |
| Angle in a semicircle (corollary of the arc theorem; the arc is a semicircle) | \( \angle \text{at the circle} = 90^\circ \) when the arc is a diameter |
| Sum of each pair of opposite angles of a cyclic quadrilateral | \( \angle A + \angle C = 180^\circ \), \( \angle B + \angle D = 180^\circ \) |
All Formulas, Grouped by Topic
The formulas of this chapter are not algebraic like speed or area formulas; they are geometric relations. The two genuinely numerical formulas are the chord–distance formula and the half-angle relation for arcs. The rest are equality results, and each is stated below exactly as the chapter presents it.
Chord Length and Distance from the Centre
From the End-of-Chapter Exercises (NCERT, p. 114–116): if the perpendicular distance of a chord from the centre is \( d \) and the radius is \( r \), then the chord is bisected by the perpendicular and the chord length is \[ \text{Chord length} = 2\sqrt{r^2 – d^2}. \]
This is a direct application of the Baudhāyana–Pythagoras theorem: the radius, half the chord and the perpendicular distance form a right triangle.
Rearranging this formula gives the other two forms the exercises ask for:
\[ d = \sqrt{r^2 – \left(\frac{\text{Chord length}}{2}\right)^2} \]
\[ r = \sqrt{d^2 + \left(\frac{\text{Chord length}}{2}\right)^2} \]
Angles Subtended by an Arc
The key relation of this sub-topic (NCERT, p. 108–110): the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the remaining part of the circle. For an arc \( AXB \) and a point \( D \) on the circle outside the arc, \[ \angle ADB = \frac{1}{2} \angle ACB \]
where \( C \) is the centre. Equivalently, the central angle is twice the inscribed angle:
\[ \angle \text{at centre} = 2 \times \angle \text{at any point on the remaining part of the circle}. \]
Angle in a Semicircle
This is the corollary that follows immediately from the arc theorem (NCERT, p. 110). When the arc is a semicircle (i.e. the chord is a diameter), the angle subtended at the centre is a straight angle, \( 180^\circ \). Therefore:
\[ \text{Angle in a semicircle} = \frac{1}{2} \times 180^\circ = 90^\circ. \]
So the angle subtended by a diameter at any point on the circle is \( 90^\circ \).
Cyclic Quadrilaterals
The central property (NCERT, p. 112–113) is that the sum of the two opposite angles of a cyclic quadrilateral is \( 180^\circ \). For a cyclic quadrilateral \( ABCD \), \[ \angle BAD + \angle BCD = 180^\circ, \qquad \angle ABC + \angle CDA = 180^\circ. \]
The converse also holds (NCERT, p. 113): if two opposite angles of a quadrilateral add up to \( 180^\circ \), then the quadrilateral is cyclic.
What Each Symbol Means
| Symbol | What it means | Unit / nature |
|---|---|---|
| \( r \) | Radius of the circle; the distance from the centre to any point on the circle | Length (cm, m, unit) |
| \( d \) | Perpendicular distance of a chord from the centre | Length (cm, m, unit) |
| \( \text{Chord length} \) | Length of the chord; \( 2\sqrt{r^2-d^2} \) | Length (cm, m, unit) |
| \( \angle ACB \) | Angle subtended by an arc at the centre \( C \) of the circle | Degrees (\( ^\circ \)) |
| \( \angle ADB \) | Angle subtended by the same arc at a point \( D \) on the remaining part of the circle | Degrees (\( ^\circ \)) |
| \( \angle A + \angle C \) | Sum of one pair of opposite angles of a cyclic quadrilateral | Degrees (\( ^\circ \)), equal to \( 180^\circ \) |
When to Use Each Formula
- Chord length from radius and distance: use \( 2\sqrt{r^2-d^2} \) when you know the radius and the perpendicular distance of the chord from the centre. Valid only when \( d \le r \); if \( d = r \) the chord is a single point of length zero.
- Distance from the centre: rearrange to \( d = \sqrt{r^2 – (\text{chord}/2)^2} \) when the radius and chord length are given.
- Radius: use \( r = \sqrt{d^2 + (\text{chord}/2)^2} \) when the chord length and its distance from the centre are given.
- Arc angle relation: use \( \angle \text{at centre} = 2 \times \angle \text{at point on circle} \) when an arc subtends an angle at the centre and you need the angle at any point on the remaining part of the circle, or vice versa.
- Angle in a semicircle: use \( 90^\circ \) whenever a diameter subtends an angle at any point on the circle.
- Cyclic quadrilateral: use \( \angle A + \angle C = 180^\circ \) and \( \angle B + \angle D = 180^\circ \) to find a missing opposite angle, and use the converse to test whether four points are concyclic.
Worked Examples
Example 1: Finding a chord length
Given: A circle has radius \( 13\ \text{cm} \) and a chord is \( 5\ \text{cm} \) away from the centre.
Find the length of the chord.
Step 1: Select the formula \( \text{Chord length} = 2\sqrt{r^2 – d^2} \).
\[ \text{Chord length} = 2\sqrt{13^2 – 5^2} = 2\sqrt{169 – 25} = 2\sqrt{144} \]
\[ = 2 \times 12 = 24\ \text{cm}. \]
Final answer: The chord length is \( 24\ \text{cm} \).
Example 2: Finding the distance of a chord from the centre
Given: The diameter of a circle is \( 26\ \text{cm} \), so the radius is \( 13\ \text{cm} \).
A chord of length \( 24\ \text{cm} \) is drawn.
Find the distance from the centre to the chord.
Step 1: Use the rearranged form \( d = \sqrt{r^2 – (\text{chord}/2)^2} \).
Here half the chord is \( 12\ \text{cm} \).
\[ d = \sqrt{13^2 – 12^2} = \sqrt{169 – 144} = \sqrt{25} = 5\ \text{cm}. \]
Final answer: The chord is \( 5\ \text{cm} \) from the centre.
Example 3: Angle in a semicircle
Given: \( AB \) is a diameter of a circle and \( C \) is any point on the circle.
Find \( \angle ACB \).
Step 1: The arc not containing \( C \) is a semicircle, which subtends \( 180^\circ \) at the centre.
\[ \angle ACB = \frac{1}{2} \times 180^\circ = 90^\circ. \]
Final answer: \( \angle ACB = 90^\circ \).
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Using the full chord length instead of half the chord inside \( r^2 – d^2 \) | The right triangle uses half the chord: \( (\text{chord}/2)^2 \). | Work backwards: does \( 2\sqrt{r^2-d^2} \) give the original chord length? |
| Forgetting the factor of 2 and writing chord length as \( \sqrt{r^2-d^2} \) | The perpendicular bisects the chord, so the full chord is twice the half-length. | Compare with the radius: the chord should be less than or equal to \( 2r \). |
| Confusing \( d = r \) with \( d = 0 \) | \( d = 0 \) means the chord passes through the centre (a diameter); \( d = r \) means the chord shrinks to a point on the circle. | Check the meaning: a chord at distance 0 from the centre is the longest chord, not the shortest. |
| Applying \( \angle \text{at centre} = 2 \times \angle \text{at circle} \) to the wrong arc | The point on the circle must lie on the remaining part of the circle, outside the arc in question. | Verify which arc is being referred to; draw the two angles and confirm they stand on the same arc. |
| Using \( \angle A + \angle C = 180^\circ \) for a quadrilateral that is not cyclic | The rule (and its converse) requires the quadrilateral to be cyclic, or one pair of opposite angles to sum to \( 180^\circ \). | Test the converse: if \( \angle A + \angle C = 180^\circ \) and \( \angle B + \angle D = 180^\circ \), the quadrilateral is cyclic. |
Frequently Asked Questions
How many circles can pass through two given points?
Infinitely many. The centres of all such circles lie on the perpendicular bisector of the line segment joining the two points (NCERT, p. 95–96). The smallest such circle has the segment as its diameter, so its radius is half the distance between the points.
How many circles can pass through three given points?
If the three points are non-collinear, exactly one circle passes through them; it is called the circumcircle of the triangle they form. If the points are collinear, no circle can pass through all three.
Why is the angle in a semicircle always 90°?
A diameter subtends a straight angle, \( 180^\circ \), at the centre. Since the angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle, the angle at the circle is half of \( 180^\circ \), which is \( 90^\circ \).
Reference: NCERT Class 9 Mathematics textbook, chapter “I’m Up and Down, and Round and Round”.
Explore Class 9 Maths Formulas
- Previous: The Dawn of Mathematics: the Human Need to Count
- Next: Measuring Space: Perimeter and Area
Related chapters:
- Orienting Yourself: The Use of Coordinates notes
- The Mathematics of Maybe: Introduction to Probability notes
- Predicting What Comes Next: Exploring Sequences and Progressions notes
Official source: download the NCERT textbook free from ncert.nic.in.