Complete theory, important theorems (Tangent Theorem, Length of Tangents Theorem), proofs, solved examples, and exercise solutions. Prepared strictly according to CBSE board exam pattern 2025-26.
A circle is a set of all points in a plane that are at a fixed distance from a fixed point. The fixed point is called the centre and the fixed distance is called the radius.
In our daily life, we see circles everywhere — wheels, coins, clocks, bangles, etc. In this chapter, we will study about tangents to a circle and their properties, which are very important for board exams.
Distance from centre to any point on the circle. Denoted by 'r'.
Line segment joining any two points on the circle.
Longest chord passing through centre. Diameter = 2 × radius.
A line that intersects the circle at two distinct points.
A line that touches the circle at exactly one point. The point is called point of contact.
A part of the circumference of a circle.
A tangent PQ at point P of a circle of radius 5 cm meets a line through centre O at a point Q such that OQ = 13 cm. Find length of PQ.
Solution: Since radius OP ⟂ tangent PQ, ∠OPQ = 90°. Using Pythagoras: PQ² = OQ² - OP² = 13² - 5² = 169 - 25 = 144 → PQ = 12 cm.
Two tangents TP and TQ are drawn from an external point T to a circle with centre O. If ∠PTQ = 80°, find ∠POQ.
Solution: In quadrilateral OQTP, ∠OQT = ∠OPT = 90°. Sum of all angles = 360°. So, ∠POQ + 90° + 80° + 90° = 360° → ∠POQ + 260° = 360° → ∠POQ = 100°.
| Theorem No. | Statement |
|---|---|
| Theorem 10.1 | The tangent at any point of a circle is perpendicular to the radius through the point of contact. |
| Theorem 10.2 | The lengths of tangents drawn from an external point to a circle are equal. |
| Corollary 1 | If two tangents are drawn from an external point, then the line joining the centre to that point bisects the angle between the tangents. |
| Corollary 2 | If two tangents are drawn from an external point, then the line joining the centre to that point bisects the angle subtended by the chord of contact at the centre. |
Question: From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the radius of the circle.
Solution: Let O be the centre. Tangent at point P, so OP ⟂ PQ. In right ΔOPQ, OQ² = OP² + PQ² ⇒ 25² = r² + 24² ⇒ 625 = r² + 576 ⇒ r² = 49 ⇒ r = 7 cm.
Question: Two tangents TP and TQ are drawn from an external point T to a circle with centre O. Prove that ∠PTQ = 2∠OPQ.
Solution: In ΔOPQ, OP = OQ ⇒ ∠OPQ = ∠OQP. Also, ∠POQ = 180° - 2∠OPQ. In quadrilateral OPTQ, ∠OPT = ∠OQT = 90°. So, ∠PTQ = 360° - (90°+90°+∠POQ) = 180° - ∠POQ = 180° - (180° - 2∠OPQ) = 2∠OPQ. Hence proved.
Question: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution: Let AB be a diameter. Tangents at A and B are drawn. Radius OA ⟂ tangent at A, so OA ⟂ line l. Similarly, OB ⟂ tangent at B, so OB ⟂ line m. Since OA and OB are in the same line (diameter), l ∥ m. Hence proved.
Question: A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Solution: Let tangents from A = AP = AS, from B = BP = BQ, from C = CQ = CR, from D = DR = DS. Adding: AP+BP+CR+DR = AS+BQ+CQ+DS ⇒ (AP+BP)+(CR+DR) = (AS+DS)+(BQ+CQ) ⇒ AB + CD = AD + BC. Hence proved.
Solution: ∠PTQ + ∠POQ = 180° ⇒ ∠POQ = 140°. In ΔOPQ, OP=OQ ⇒ ∠OPQ = ∠OQP = (180°-140°)/2 = 20°.
Solution: Let AB be chord. Tangents at A and B meet at P. Join OA, OB. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. ∠OAB = ∠OBA (OA=OB). Hence ∠PAB = ∠PBA. Hence proved.
Solution: Let sides touch at D, E, F. Then AD = AF, BD = BE, CE = CF. Adding gives the required property.