Circles — Class 10 Mathematics
1. Tangent to a Circle
A TANGENT is a line that touches the circle at EXACTLY ONE POINT. That point is the POINT OF CONTACT (or point of tangency). 'A tangent TOUCHES the circle. A secant CUTS through the circle at two points. A chord is a segment INSIDE the circle connecting two points on it.'
Key Theorem — Radius and Tangent
The tangent at ANY point of a circle is PERPENDICULAR to the RADIUS through the point of contact. If PT is a tangent to the circle with centre O at point P, then OP ⟂ PT. 'This is the FUNDAMENTAL property of tangents. Every tangent problem uses this theorem.'
2. Number of Tangents from a Point
| Position of Point | Number of Tangents | Explanation |
|---|---|---|
| Inside the circle | 0 | Any line through an interior point cuts the circle at TWO points (secant) |
| ON the circle | 1 | Exactly ONE tangent — perpendicular to the radius at that point |
| Outside the circle | 2 | TWO tangents can be drawn. They are EQUAL in LENGTH. |
Theorem — Tangents from External Point: The LENGTHS of tangents drawn from an EXTERNAL point to a circle are EQUAL. If P is an external point, PA and PB are tangents (A and B are points of contact), then PA = PB.
Proof
Join OA, OB, and OP. OA = OB (radii). ∠OAP = ∠OBP = 90° (radius ⟂ tangent). OP = OP (common). By RHS → ΔOAP ≅ ΔOBP → PA = PB (CPCT). Also, ∠APO = ∠BPO — OP bisects the angle between the tangents.
3. Worked Examples
Example 1 — Tangent and Radius: A tangent PQ touches a circle with centre O at P. OQ = 15 cm and PQ = 12 cm. Find the radius. OP ⟂ PQ. In right ΔOPQ: OP² + PQ² = OQ² → r² + 144 = 225 → r² = 81 → r = 9 cm.
Example 2 — Tangents from External Point: PT and PS are tangents from point P to circle with centre O. PT = 24 cm, radius = 7 cm. Find OP. OP² = PT² + OT² = 24² + 7² = 576 + 49 = 625. OP = 25 cm.
Example 3 — Angle Between Tangents: PT and PS are tangents from P. ∠TPS = 60°. Find ∠TOS (where O is centre). In quadrilateral PTOS: ∠OTP = ∠OSP = 90° (radius ⟂ tangent). ∠TPS = 60° (given). Sum of angles in quadrilateral = 360°. ∠TOS = 360° − 90° − 90° − 60° = 120°. 'The angle between the two radii to the points of contact is SUPPLEMENTARY to the angle between the tangents: ∠TOS + ∠TPS = 180°.'
Example 4 — Circle Touching a Line: A circle of radius 5 cm touches a line at P. A point Q on the line is 13 cm from the centre O. Find PQ. OP = 5 cm (radius). OQ = 13 cm. Since OP ⟂ line: PQ² = OQ² − OP² = 169 − 25 = 144. PQ = 12 cm.
Example 5 — Variable Tangent Length: The length of a tangent from a point at distance d from the centre is √(d²−r²). For the tangent to EXIST: d ≥ r. If d = r (point on circle): tangent length = 0 (it's the point itself). If d < r (point inside): no tangent exists.
4. Construction of Tangents (Practical Geometry)
To draw a tangent at a point ON the circle
- Draw radius to the point of contact.
- Construct a line PERPENDICULAR to this radius at the point.
- This line is the required tangent.
To draw tangents from an EXTERNAL point
- Join the external point P to centre O.
- Bisect OP. Let M be the midpoint.
- With M as centre and MP as radius, draw a circle. It intersects the original circle at A and B.
- Join PA and PB. These are the required tangents. 'This construction uses the fact that the angle in a semicircle is 90°. Points A and B are where the semicircle on OP (as diameter) meets the original circle — and PA and PB are perpendicular to OA and OB respectively.'
5. Tangent and Secant
A SECANT intersects the circle at TWO points. A TANGENT touches at ONE point. 'A tangent is the LIMITING CASE of a secant — when the two points of intersection COINCIDE.'
6. Common Mistakes
- Assuming a tangent passes through the centre: A tangent NEVER passes through the centre (except at... no, never). The radius TO the point of contact is perpendicular to the tangent.
- Thinking a line touching the circle is automatically tangent: It must touch at EXACTLY ONE point.
- Forgetting that tangents from an external point are EQUAL: This is the most commonly used property in proofs.
- Measure zero-length as "no tangent": From a point ON the circle, the "tangent length" is 0, but the tangent LINE exists.
7. AP Exam Focus
| Topic | Marks |
|---|---|
| Radius ⟂ tangent property | 3-4 |
| Tangents from external point (equality proof) | 4-5 |
| Length of tangent calculation | 3-4 |
| Construction of tangents | 4-5 |
Quick Self-Test
- How many tangents can be drawn from a point outside a circle? (Answer: 2.)
- PA and PB are tangents from P. PA = 8 cm. Find PB. (Answer: 8 cm — tangents from same external point are equal.)
- A tangent at P touches circle (O,5cm). Q is a point on the tangent, OQ = 13 cm. Find PQ. (Answer: √(169−25) = √144 = 12 cm.)
- What is the angle between a radius and the tangent at its endpoint? (Answer: 90°.)
- Distance from centre to external point = 10 cm, radius = 6 cm. Tangent length? (Answer: √(100−36) = √64 = 8 cm.)
More Worked Examples
Example 5 — Circle Touching Two Lines: A circle touches two perpendicular lines. The centre is at distance 5 cm from each line. Find the radius. Since the circle touches both lines, and the centre is equidistant (5 cm) from both, the radius = 5 cm. The centre lies on the angle bisector of the two lines.
Example 6 — Circle Inscribed in a Triangle: A circle with centre I is inscribed in ΔABC, touching BC, CA, AB at D, E, F. Prove that BD = s−b, where s is the semi-perimeter. Let BD = x, DC = y, and... actually: AB = c, BC = a, CA = b. Tangents from B: BF = BD (both from B). Similarly, CE = CD, AF = AE. Let BD=BF=x, CD=CE=y, AE=AF=z. Then: x+y=a, y+z=b, z+x=c. Solving: x = (a+c−b)/2 = s−b. 'This is a classic result — the lengths from vertices to points of tangency of the incircle are directly expressible in terms of the semi-perimeter.'
Example 7 — Proof of Tangent Property in Quadrilateral: A circle touches all four sides of a quadrilateral ABCD. Prove AB + CD = AD + BC. Let the circle touch AB, BC, CD, DA at P, Q, R, S respectively. Tangents from A: AP = AS. From B: BP = BQ. From C: CQ = CR. From D: DR = DS. Adding: AB+CD = (AP+BP)+(CR+DR) = (AS+BQ)+(CQ+DS) = (AS+DS)+(BQ+CQ) = AD+BC. 'This is Pitot's Theorem — a bicentric quadrilateral has equal sums of opposite sides.'
Theorem Proof — Length of Tangent from External Point
Given: Circle with centre O. P is an external point. PT is tangent at T. To Prove: PT = √(OP²−OT²). Construction: Join OT, OP. Proof: ∠OTP = 90° (radius ⟂ tangent). In right ΔOTP: OP² = OT² + PT² → PT² = OP²−OT² → PT = √(d²−r²). Corollary: PT = √(d²−r²). For the tangent to exist, d ≥ r.
Circles in Daily Life and AP Context
In Andhra Pradesh, the famous Lepakshi Temple features a hanging pillar — a marvel of balance and geometry. The arches of the Kakatiya gateways (Warangal) are segments of circles. 'Understanding circles is understanding the geometry of arches, domes, and wheels — the foundations of architecture and engineering that surround us in Andhra's magnificent temples and monuments.'
Common Exam Pitfalls
- Writing PA = PB without justification — ALWAYS state 'Tangents from an external point are EQUAL.'
- Using the wrong right triangle — The right angle is at the POINT OF CONTACT, not at the centre.
- Forgetting to state that OP bisects ∠APB when two tangents are drawn — this follows from the congruence of ΔOAP and ΔOBP.
AP Exam Question Types for Circles
| Type | What to Do | Marks |
|---|---|---|
| Find tangent length | Use PT = √(d²−r²) | 3-4 |
| Prove PA = PB | Full proof with RHS congruence | 4-5 |
| Construction | Draw tangents from external point | 4-5 |
| Angle between tangents | Use quadrilateral angle sum (360° − 90° − 90° − angle) | 3-4 |
| Application to quadrilaterals | Use equal tangent property repeatedly | 4-5 |
Key Points to Remember
- The tangent is ALWAYS perpendicular to the radius at the point of contact.
- From any external point, exactly two tangents can be drawn, and they are EQUAL in length.
- The line joining the external point to the centre BISECTS the angle between the tangents.
- The angle between the two tangents and the angle subtended by the chord of contact at the centre are SUPPLEMENTARY: ∠TOS + ∠TPS = 180°.
- Circle geometry is PURE deduction — every step must be justified by a theorem or property.
