Circles studies what happens when a straight line and a circle lie in the same plane. A line can miss the circle, cut it at two points as a secant, or touch it at exactly one point as a tangent. The chapter proves two theorems: the tangent at any point is perpendicular to the radius through the point of contact, and the two tangents from an external point are equal. With Pythagoras, these results find tangent lengths, angles between tangents, chords of concentric circles and the sides of quadrilaterals and triangles drawn around a circle.
Key Concepts
1. A Line and a Circle: Three Possible Positions
From Class 9, a circle is the collection of all points in a plane that are at a constant distance (the radius) from a fixed point (the centre). This chapter asks how a straight line PQ can sit relative to a circle in the same plane. Only three things can happen:
| Position | Common points | Name |
|---|---|---|
| The line stays outside the circle | None | Non-intersecting line |
| The line cuts the circle at A and B | Two | Secant |
| The line touches the circle at A only | Exactly one | Tangent |
There is no fourth possibility. Keep secant and chord separate: the secant is the whole line PQ; the chord is only the segment AB between the two meeting points.
Tangents in daily life. The rope on either side of a pulley over a well, thought of as a ray, is like a tangent to the circle of the pulley. A wheel moves along a line that is a tangent to the circle of the wheel, and the spoke pointing to the ground at the point of contact looks perpendicular to the road. That second picture is Theorem 10.1.
“Tangent” comes from the Latin word tangere, which means “to touch”. It was introduced by the Danish mathematician Thomas Fineke in 1583.
2. The Tangent as a Special Secant
Activity 1. A straight wire AB is fixed at a point P of a circular wire so that it can turn about P. In most positions it cuts the circle at P and at a second point. As it turns, the second point slides closer to P, and in exactly one position it reaches P: the wire then meets the circle at P only. That position is the tangent at P. So a tangent exists at P, and there is only one tangent at a point of the circle.
Activity 2. Draw a secant and lines parallel to it on both sides. Moving away from the centre, the chords they cut get shorter. On each side, one parallel line cuts a chord of length zero and touches the circle. So there cannot be more than two tangents parallel to a given secant, one on each side.
Both activities lead to the NCERT statement: the tangent to a circle is a special case of the secant, when the two end points of its corresponding chord coincide.
The common point of the tangent and the circle is the point of contact, and the tangent is said to touch the circle there. Since there is one tangent at each of the infinitely many points of a circle, a circle has infinitely many tangents (NCERT Exercise 10.1, Question 1).
3. Theorem 10.1: The Tangent Is Perpendicular to the Radius
Statement. The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: a circle with centre O and a tangent XY to the circle at a point P.
To prove: OP ⊥ XY.
Proof.
- Take any point Q on XY other than P, and join OQ.
- Q must lie outside the circle. If Q were inside the circle, the line XY would cut the circle at two points and would be a secant, not a tangent.
- Since Q is outside the circle, OQ is longer than the radius OP. That is, OQ > OP.
- This is true for every point Q on XY except P. So OP is the shortest of all the distances from O to the points of the line XY.
- The shortest distance from a point to a line is along the perpendicular to the line. Therefore OP ⊥ XY.
Remarks. At any point on a circle there can be one and only one tangent, since only one line is perpendicular to OP at P. The line containing the radius through the point of contact is sometimes called the normal to the circle at that point.
How the theorem is used. Whenever a radius meets a point of contact, mark a right angle there. The right triangle it creates usually leads straight to Pythagoras theorem.
4. Number of Tangents from a Point and the Length of a Tangent
NCERT Activity 3 tries drawing tangents from a point P in three positions:
| Where P lies | Tangents through P | Reason |
|---|---|---|
| Inside the circle | None | Every line through an inside point cuts the circle in two points |
| On the circle | One and only one | Theorem 10.1: only one line is perpendicular to OP at P |
| Outside the circle | Exactly two | One tangent touches on each side of the line OP |
Length of a tangent. If the tangents from an outside point P touch the circle at T1 and T2, the segments PT1 and PT2 are the lengths of the tangents from P: the segment from the external point to the point of contact.
The working formula. Let the circle have centre O and radius r, and let a tangent from P touch it at T. Join OT and OP. By Theorem 10.1, ∠OTP = 90°, so triangle OTP is right-angled at T with hypotenuse OP.
- PT2 = OP2 − r2 (length of the tangent)
- r2 = OP2 − PT2 (radius)
- OP2 = r2 + PT2 (distance of P from the centre)
OP is the hypotenuse, so it is always the longest of the three.
| Radius r | Distance OP | Tangent length PT | Working |
|---|---|---|---|
| 7 cm | 25 cm | 24 cm | √(625 − 49) = √576 |
| 3 cm | 5 cm | 4 cm | √(25 − 9) = √16 |
| 5 cm | 13 cm | 12 cm | √(169 − 25) = √144 |
5. Theorem 10.2: Tangents from an External Point Are Equal
Statement. The lengths of tangents drawn from an external point to a circle are equal.
Given: a circle with centre O, a point P outside the circle, and two tangents PQ and PR to the circle from P, touching it at Q and R.
To prove: PQ = PR.
Construction: join OP, OQ and OR.
Proof.
- ∠OQP = 90° and ∠ORP = 90°, because each is the angle between a radius and the tangent at its point of contact (Theorem 10.1).
- In right triangles OQP and ORP: OQ = OR (radii of the same circle) and OP = OP (common hypotenuse).
- So ΔOQP ≅ ΔORP by the RHS congruence rule.
- Therefore PQ = PR (CPCT).
Second proof, by Pythagoras. PQ2 = OP2 − OQ2 = OP2 − OR2 = PR2, since OQ = OR. So PQ = PR.
Extra results from the same congruence. Since ΔOQP ≅ ΔORP, the matching angles are also equal:
- ∠OPQ = ∠OPR. So OP is the bisector of ∠QPR: the centre lies on the bisector of the angle between the two tangents.
- ∠POQ = ∠POR. So OP also bisects the angle QOR at the centre.
6. Angles Made by Two Tangents
Let TP and TQ be tangents from an external point T to a circle with centre O, touching it at P and Q. Three angle facts follow.
(a) The angle between the tangents and the angle at the centre are supplementary
In quadrilateral OPTQ, ∠OPT = ∠OQT = 90° (Theorem 10.1). The angles of a quadrilateral add up to 360°, so
∠PTQ + ∠POQ = 360° − 90° − 90° = 180°.
In words: the angle between the two tangents drawn from an external point is supplementary to the angle subtended at the centre by the line segment joining the points of contact. This is NCERT Exercise 10.2, Question 10.
| ∠PTQ (between tangents) | ∠POQ (at the centre) | ∠POT = ∠QOT | ∠PTO = ∠QTO |
|---|---|---|---|
| 70° | 110° | 55° | 35° |
| 80° | 100° | 50° | 40° |
| 90° | 90° | 45° | 45° |
(b) ∠PTQ = 2∠OPQ
This is NCERT Example 2, proved in the Solved Examples. TP = TQ makes triangle TPQ isosceles, and ∠OPT = 90° turns its base angle into half of ∠PTQ.
(c) OT is the perpendicular bisector of the chord of contact PQ
Call PQ, the segment joining the points of contact, the chord of contact. Triangle TPQ is isosceles and TO bisects its vertex angle, so TO ⊥ PQ and TO bisects PQ. NCERT Example 3 uses this.
7. Concentric Circles and a Chord That Touches the Inner Circle
Two circles with the same centre are concentric. If a chord AB of the larger circle touches the smaller circle at P, then OP ⊥ AB (Theorem 10.1), and the perpendicular from the centre to a chord bisects it, so AP = PB. This is NCERT Example 1, proved in full in the Solved Examples.
Length of that chord. Let the radii be R (larger) and r (smaller). In right triangle OPA, OA = R and OP = r, so AP = √(R2 − r2) and
AB = 2√(R2 − r2)
With R = 5 cm and r = 3 cm, AB = 2√(25 − 9) = 2 × 4 = 8 cm (NCERT Exercise 10.2, Question 7).
8. Polygons Drawn Around a Circle
A polygon circumscribes a circle when every side touches the circle. Each vertex is then an external point, and by Theorem 10.2 its two tangent segments are equal.
Quadrilateral circumscribing a circle
Let a circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S. Tangents from each vertex give AP = AS, BP = BQ, CR = CQ and DR = DS. Adding:
AB + CD = (AP + PB) + (CR + RD) = AS + BQ + CQ + DS = (AS + DS) + (BQ + CQ) = AD + BC.
So the sums of opposite sides are equal. If AB = 6 cm, BC = 7 cm and CD = 4 cm, then AD = 6 + 4 − 7 = 3 cm.
Parallelogram circumscribing a circle. In a parallelogram AB = CD and AD = BC. Putting these into AB + CD = AD + BC gives 2AB = 2AD, so AB = AD. A parallelogram with two adjacent sides equal has all four sides equal, so it is a rhombus (NCERT Exercise 10.2, Question 11). A rectangle that is not a square cannot circumscribe a circle.
Angles at the centre. Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre: ∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180° (NCERT Exercise 10.2, Question 13, proved below).
Triangle circumscribing a circle
Let a circle touch BC at D, CA at E and AB at F. Tangents from each vertex give
AE = AF, BD = BF, CD = CE.
Call these x, y and z: AF = AE = x, BF = BD = y, CD = CE = z. Then AB = x + y, BC = y + z and CA = z + x. Adding, the perimeter is 2(x + y + z). If s is half the perimeter, then
x = s − BC, y = s − CA, z = s − AB.
Area and the radius. Join the centre O to A, B and C. The triangle splits into triangles OBC, OCA and OAB. Each has height r (the radius is perpendicular to the side it touches), so area of ABC = ½ r (BC + CA + AB) = r × s. NCERT Exercise 10.2, Question 12 uses this area, together with Heron’s formula from Class 9, to find two unknown sides.
Formula and Theorem Sheet
| Result | Statement or formula |
|---|---|
| Positions of a line | 0 common points: non-intersecting; 2: secant; 1: tangent |
| Parallel tangents | At most two tangents are parallel to a given secant |
| Theorem 10.1 | Tangent at any point ⊥ radius through the point of contact; exactly one tangent at each point |
| Tangents through a point | Inside: 0; on the circle: 1; outside: 2 |
| Length of tangent | PT2 = OP2 − r2 |
| Theorem 10.2 | Tangents from an external point are equal: PQ = PR; OP bisects ∠QPR |
| Two tangents TP, TQ | ∠PTQ + ∠POQ = 180°; ∠PTQ = 2∠OPQ; OT is the perpendicular bisector of PQ |
| Concentric circles | Chord of larger circle touching smaller is bisected at the point of contact; length 2√(R2 − r2) |
| Circumscribing quadrilateral | AB + CD = AD + BC; opposite sides subtend supplementary angles at the centre |
| Circumscribing parallelogram | Must be a rhombus |
| Parallel tangents cut by a third tangent at A, B | ∠AOB = 90° |
| Circumscribing triangle | Tangent segments s − a, s − b, s − c; area = r × s |
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Important Definitions
- Circle: the collection of all points in a plane which are at a constant distance (radius) from a fixed point (centre).
- Non-intersecting line: a line that has no common point with the circle.
- Secant: a line that intersects a circle in two distinct points.
- Chord: the line segment joining the two points where a secant meets the circle.
- Tangent: a line that intersects the circle at only one point. It is a special case of a secant when the two end points of its corresponding chord coincide.
- Point of contact: the common point of a tangent and the circle. The tangent is said to touch the circle at this point.
- Normal: the line containing the radius through the point of contact. It is perpendicular to the tangent at that point.
- Length of a tangent: the length of the segment of the tangent from the external point to the point of contact with the circle.
- Concentric circles: circles in the same plane with the same centre and different radii.
- Circumscribing polygon: a polygon all of whose sides touch a circle. The circle is then said to be inscribed in the polygon.
Solved Examples (NCERT-Based)
Example 1: Length of a tangent (NCERT Exercise 10.1, Q3)
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is: (A) 12 cm (B) 13 cm (C) 8.5 cm (D) √119 cm.
Solution: OP is the radius to the point of contact P, so by Theorem 10.1, ∠OPQ = 90°. In right triangle OPQ, the hypotenuse is OQ.
PQ2 = OQ2 − OP2 = 122 − 52 = 144 − 25 = 119.
PQ = √119 cm. Answer: (D). Option (B) is the trap: 13 comes from adding the squares, which treats PQ as the hypotenuse.
Example 2: Finding the radius (NCERT Exercise 10.2, Q1 and Q6)
(i) 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. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm.
(ii) The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution: Let the tangent touch the circle at P. By Theorem 10.1, ∠OPQ = 90°, so OQ is the hypotenuse.
(i) r2 = OQ2 − PQ2 = 625 − 576 = 49, so r = 7 cm. Answer: (A).
(ii) Here the external point is A, with OA = 5 cm and the tangent length 4 cm. r2 = OA2 − 42 = 25 − 16 = 9, so r = 3 cm.
Example 3: Angle between tangents from the angle at the centre (NCERT Exercise 10.2, Q2)
If TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to (A) 60° (B) 70° (C) 80° (D) 90°.
Solution: In quadrilateral OPTQ, ∠OPT = ∠OQT = 90° (Theorem 10.1). The angle sum of a quadrilateral is 360°.
∠PTQ = 360° − 90° − 90° − 110° = 70°. Answer: (B).
Example 4: Angle at the centre from the angle between tangents (NCERT Exercise 10.2, Q3)
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to (A) 50° (B) 60° (C) 70° (D) 80°.
Solution: OP bisects ∠APB (the centre lies on the bisector of the angle between the tangents), so ∠OPA = 80°/2 = 40°.
In triangle OAP, ∠OAP = 90° (Theorem 10.1). So ∠POA = 180° − 90° − 40° = 50°. Answer: (A).
Example 5: Chord of the larger of two concentric circles (NCERT Example 1)
Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Given: two concentric circles C1 and C2 with centre O, and a chord AB of the larger circle C1 which touches the smaller circle C2 at P.
To prove: AP = BP.
Proof: Join OP. AB is a tangent to C2 at P and OP is a radius of C2. So by Theorem 10.1, OP ⊥ AB.
Now AB is a chord of C1 and OP ⊥ AB. The perpendicular from the centre of a circle to a chord bisects the chord. So OP bisects AB, that is, AP = BP.
Example 6: Length of that chord (NCERT Exercise 10.2, Q7)
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution: Let the chord AB of the larger circle touch the smaller circle at P, and let O be the centre. By Theorem 10.1, OP ⊥ AB, and by Example 5, P is the midpoint of AB.
In right triangle OPA: OA = 5 cm (radius of the larger circle), OP = 3 cm (radius of the smaller circle).
AP = √(OA2 − OP2) = √(25 − 9) = √16 = 4 cm.
AB = 2 × AP = 8 cm.
Example 7: ∠PTQ = 2∠OPQ (NCERT Example 2)
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.
Proof: Let ∠PTQ = θ.
By Theorem 10.2, TP = TQ, so triangle TPQ is isosceles and its base angles are equal:
∠TPQ = ∠TQP = ½(180° − θ) = 90° − θ/2.
By Theorem 10.1, ∠OPT = 90°. So
∠OPQ = ∠OPT − ∠TPQ = 90° − (90° − θ/2) = θ/2 = ½∠PTQ.
This gives ∠PTQ = 2∠OPQ.
Example 8: Tangents at the ends of a chord (NCERT Example 3)
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.
Solution (similar triangles): Join OT and let it meet PQ at R. Triangle TPQ is isosceles (TP = TQ) and TO is the bisector of ∠PTQ, so OT ⊥ PQ and OT bisects PQ. Hence PR = RQ = 4 cm.
In right triangle ORP: OR = √(OP2 − PR2) = √(25 − 16) = 3 cm.
Now ∠TPR + ∠RPO = 90° (because ∠TPO = 90°, Theorem 10.1), and ∠TPR + ∠PTR = 90° (angle sum of triangle TRP, which is right-angled at R). So ∠RPO = ∠PTR.
Therefore right triangle TRP is similar to right triangle PRO by AA similarity. This gives
TP/PO = RP/RO, that is TP/5 = 4/3, so TP = 20/3 cm.
Second method (Pythagoras): Let TP = x and TR = y.
From right triangle PRT: x2 = y2 + 16 … (1)
From right triangle OPT: x2 + 52 = (y + 3)2 … (2)
Subtracting (1) from (2): 25 = 6y + 9 − 16 = 6y − 7, so y = 32/6 = 16/3.
Then x2 = (16/3)2 + 16 = 256/9 + 144/9 = 400/9, so x = 20/3 cm. Both methods agree.
Example 9: Tangents at the ends of a diameter, and the perpendicular at the point of contact (NCERT Exercise 10.2, Q4 and Q5)
(i) Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
(ii) Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Proof of (i): Let AB be a diameter of a circle with centre O. Let PQ be the tangent at A and RS the tangent at B. By Theorem 10.1, OA ⊥ PQ and OB ⊥ RS. Taking P and S on opposite sides of AB, ∠PAB = 90° and ∠ABS = 90°. These are alternate angles made by the transversal AB with the lines PQ and RS, and they are equal. Hence PQ ∥ RS.
Proof of (ii): Let AB be the tangent at P to a circle with centre O. By Theorem 10.1, OP ⊥ AB. At a point of a line only one perpendicular can be drawn in the plane, so the perpendicular to AB at P is the line OP, which passes through the centre O.
Example 10: Quadrilateral circumscribing a circle (NCERT Exercise 10.2, Q8 and Q11)
(i) A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
(ii) Prove that the parallelogram circumscribing a circle is a rhombus.
Proof of (i): Let the circle touch AB, BC, CD and DA at P, Q, R and S. Tangents from an external point are equal (Theorem 10.2), so
AP = AS, BP = BQ, CR = CQ, DR = DS.
Adding these four equations: AP + BP + CR + DR = AS + BQ + CQ + DS.
Grouping: (AP + PB) + (CR + RD) = (AS + SD) + (BQ + QC), that is AB + CD = AD + BC.
Proof of (ii): Let ABCD be a parallelogram circumscribing a circle. By part (i), AB + CD = AD + BC. Opposite sides of a parallelogram are equal, so CD = AB and BC = AD. Then 2AB = 2AD, so AB = AD. So AB = BC = CD = DA, and ABCD is a rhombus.
Example 11: Parallel tangents and a third tangent (NCERT Exercise 10.2, Q9)
XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠AOB = 90°.
Proof: Let XY touch the circle at P and X′Y′ touch it at Q. Join OA, OB, OC, OP and OQ.
From A, the tangents AP and AC are drawn, so OA bisects ∠PAC (Theorem 10.2, remark): ∠OAC = ½∠PAC.
From B, the tangents BQ and BC are drawn, so OB bisects ∠QBC: ∠OBC = ½∠QBC.
XY ∥ X′Y′ and AB is a transversal, so ∠PAC and ∠QBC are interior angles on the same side of the transversal: ∠PAC + ∠QBC = 180°.
So ∠OAB + ∠OBA = ½(∠PAC + ∠QBC) = 90°.
In triangle AOB, ∠AOB = 180° − 90° = 90°.
Example 12: Triangle circumscribing a circle (NCERT Exercise 10.2, Q12)
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Solution: Let the circle with centre O touch AB at F and AC at E. Tangents from each vertex are equal:
BF = BD = 8 cm, CE = CD = 6 cm, AF = AE = x cm (say).
So AB = x + 8, AC = x + 6 and BC = 8 + 6 = 14 cm.
Step 1: area by splitting at the centre O (main method). Join OA, OB and OC. OD, OE and OF are radii of 4 cm, each perpendicular to its side (Theorem 10.1), so each is the height of one of the three small triangles. So
ar(ABC) = ar(OBC) + ar(OCA) + ar(OAB) = ½ × 4 × [14 + (x + 6) + (x + 8)] = 2(2x + 28) = 4(x + 14).
Step 2: a second expression for the same area. The area in Step 1 still contains the unknown x, so one more equation is needed. The standard way is Heron’s formula from Class 9, area = √[s(s − a)(s − b)(s − c)]. Semi-perimeter s = (14 + x + 6 + x + 8)/2 = x + 14. Then s − BC = x, s − CA = 8 and s − AB = 6, so
ar(ABC) = √[(x + 14)(x)(8)(6)] = √[48x(x + 14)].
Step 3: equate and solve. √[48x(x + 14)] = 4(x + 14). Squaring: 48x(x + 14) = 16(x + 14)2. Divide by 16(x + 14), which is not zero: 3x = x + 14, so x = 7.
AB = 7 + 8 = 15 cm and AC = 7 + 6 = 13 cm.
Check: sides 15, 14 and 13 give s = 21 and area √(21 × 6 × 7 × 8) = √7056 = 84 cm2, and r × s = 4 × 21 = 84 cm2. Both agree.
Example 13: Opposite sides subtend supplementary angles (NCERT Exercise 10.2, Q13)
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Proof: Let ABCD circumscribe a circle with centre O, touching AB, BC, CD and DA at P, Q, R and S. Join OA, OB, OC, OD and OP, OQ, OR, OS.
In right triangles OAP and OAS: OP = OS (radii), OA is common and ∠OPA = ∠OSA = 90°. So ΔOAP ≅ ΔOAS (RHS), giving ∠AOP = ∠AOS. Call this angle 1.
In the same way: ∠BOP = ∠BOQ (angle 2), ∠COQ = ∠COR (angle 3), ∠DOR = ∠DOS (angle 4).
The eight angles round O add up to 360°: 2(angle 1 + angle 2 + angle 3 + angle 4) = 360°, so angle 1 + angle 2 + angle 3 + angle 4 = 180°.
Now ∠AOB = angle 1 + angle 2 and ∠COD = angle 3 + angle 4. So ∠AOB + ∠COD = 180°. Similarly ∠BOC + ∠AOD = (angle 2 + angle 3) + (angle 4 + angle 1) = 180°.
Competency-Based Questions (with answers)
1. Case-based: The gate and the circular park
A circular park has radius 9 m and centre O. A gate G stands 15 m from O. Two straight fences are laid from G so that each just touches the park boundary, one at A and one at B.
(a) Find the length of each fence. ∠OAG = 90° (Theorem 10.1). GA = √(152 − 92) = √(225 − 81) = √144 = 12 m. By Theorem 10.2, GB = 12 m too.
(b) Find the area of the region OAGB enclosed by the two fences and the radii OA, OB. It is made of two congruent right triangles OAG and OBG. Area = 2 × ½ × 9 × 12 = 108 m2.
(c) Find the straight distance AB. OG is the perpendicular bisector of AB; let it meet AB at M. Area of triangle OAG = ½ × OA × AG = ½ × OG × AM, so 9 × 12 = 15 × AM, giving AM = 7.2 m. AB = 2 × 7.2 = 14.4 m.
2. Case-based: The fountain in the triangular lawn
A triangular lawn ABC has AB = 12 m, BC = 8 m and CA = 10 m. A circular fountain is built so that its edge touches all three sides: BC at D, CA at E and AB at F.
(a) Which pairs of segments are equal? AF = AE, BF = BD and CD = CE, because tangents from the same external point are equal.
(b) Find AF, BD and CE. Half the perimeter is s = (12 + 8 + 10)/2 = 15. AF = s − BC = 7 m, BD = s − CA = 5 m, CE = s − AB = 3 m.
3. Source-based: The pulley and the wheel
Read the passage adapted from the textbook: “You might have seen a pulley fitted over a well which is used in taking out water from the well. Here the rope on both sides of the pulley, if considered as a ray, is like a tangent to the circle representing the pulley.” The textbook also notes that a wheel moves along a line which is a tangent to the circle representing the wheel, and the radius through the point of contact with the ground appears to be at right angles to it.
(a) Which theorem does the wheel observation illustrate? Theorem 10.1: the tangent at any point of a circle is perpendicular to the radius through the point of contact.
(b) A wheel of radius 35 cm rests on the road at point P. A pebble on the road is 84 cm from P. How far is the pebble from the centre of the wheel? The radius to P is perpendicular to the road, so distance = √(352 + 842) = √(1225 + 7056) = √8281 = 91 cm.
4. Assertion-Reason
Assertion (A): No tangent can be drawn to a circle from a point inside it.
Reason (R): Every line through a point inside a circle intersects the circle in two points.
Options: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). Both are true. Since every line through an inside point is a secant, none of them can be a tangent, which is exactly why A holds.
5. Assertion-Reason
Assertion (A): The tangents drawn at the ends of a diameter of a circle are parallel.
Reason (R): The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Options: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). Both tangents are perpendicular to the same diameter by R, and two lines perpendicular to the same line are parallel.
6. Assertion-Reason
Assertion (A): If PA and PB are tangents to a circle with centre O and ∠APB = 60°, then ∠AOB = 120°.
Reason (R): The angle between two tangents from an external point and the angle subtended at the centre by the segment joining the points of contact are complementary.
Options: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (c). A is true: ∠AOB = 360° − 90° − 90° − 60° = 120°. R is false: the two angles are supplementary (sum 180°), not complementary (sum 90°).
7. Error analysis: The wrong tangent length
A student is told that a point P is 10 cm from the centre of a circle of radius 6 cm, and writes: “Length of tangent = √(102 + 62) = √136 cm.” Find the error and correct it.
Answer: The student treated the tangent as the hypotenuse. The right angle is at the point of contact T (Theorem 10.1), so OP = 10 cm is the hypotenuse. PT = √(102 − 62) = √64 = 8 cm. Quick check: a tangent length must be shorter than the distance OP.
8. Competency MCQ: The incircle of a right triangle
A circle touches all three sides of triangle ABC, right-angled at B, with AB = 6 cm and BC = 8 cm. Its radius is: (a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm.
Answer: (b). AC = √(36 + 64) = 10 cm. The radii to the points of contact on AB and BC, with the legs, form a square of side r at B. Tangent lengths from A and C are 6 − r and 8 − r, and they add up to AC: (6 − r) + (8 − r) = 10, so r = 2 cm.
Important Questions for Board Exams
1-Mark Questions
- How many tangents can a circle have? Infinitely many, one at each point of the circle.
- A point P is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from P. √(169 − 25) = √144 = 12 cm.
- PA and PB are tangents from P to a circle with centre O. If ∠APB = 50°, find ∠AOB. ∠AOB = 180° − 50° = 130°.
- Fill in the blank: A circle can have ______ parallel tangents at the most. Two.
2-Mark Questions
- Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord. Let the tangents at the ends P and Q of chord PQ meet at T. By Theorem 10.2, TP = TQ, so triangle TPQ is isosceles. Angles opposite equal sides are equal, so ∠TPQ = ∠TQP.
- Two concentric circles have radii 13 cm and 5 cm. Find the length of the chord of the larger circle that touches the smaller circle. Half chord = √(169 − 25) = 12 cm, so the chord is 24 cm (the point of contact bisects it).
- PA and PB are tangents from P to a circle, and ∠APB = 60°. Prove that triangle PAB is equilateral. PA = PB (Theorem 10.2), so ∠PAB = ∠PBA = (180° − 60°)/2 = 60°. All three angles are 60°, so triangle PAB is equilateral.
- A quadrilateral ABCD circumscribes a circle. AB = 8 cm, BC = 10 cm and CD = 9 cm. Find AD. AB + CD = AD + BC, so 17 = AD + 10 and AD = 7 cm.
3-Mark Questions
- Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Let XY touch the circle with centre O at P. Take any point Q on XY other than P and join OQ. Q lies outside the circle, since otherwise XY would be a secant. So OQ > OP. This holds for every point of XY except P, so OP is the shortest distance from O to XY. The shortest segment from a point to a line is perpendicular to it, so OP ⊥ XY.
- A circle is inscribed in triangle ABC with AB = 10 cm, BC = 8 cm and CA = 12 cm. It touches AB at P, BC at Q and CA at R. Find AP, BQ and CR. Let AP = AR = x, BP = BQ = y, CQ = CR = z. Then x + y = 10, y + z = 8, z + x = 12. Adding, x + y + z = 15. So x = 15 − 8 = 7, y = 15 − 12 = 3, z = 15 − 10 = 5. AP = 7 cm, BQ = 3 cm, CR = 5 cm.
- A circle touches the side BC of triangle ABC at P and touches AB and AC produced at Q and R respectively. Prove that AQ = ½(perimeter of triangle ABC). Tangents from A, B and C give AQ = AR, BQ = BP and CP = CR. Perimeter = AB + BC + CA = AB + BP + PC + CA = AB + BQ + CR + CA = AQ + AR = 2AQ. So AQ = ½ perimeter.
5-Mark Questions
- From a point T outside a circle of centre O and radius 6 cm, two tangents TP and TQ are drawn. OT = 10 cm. Find (i) TP, (ii) the area of quadrilateral OPTQ, (iii) the length of the chord PQ. (i) ∠OPT = 90°, so TP = √(100 − 36) = 8 cm. (ii) OPTQ is made of two congruent right triangles, so area = 2 × ½ × 6 × 8 = 48 cm2. (iii) OT is the perpendicular bisector of PQ; let it meet PQ at R. Area of triangle OPT = ½ × OP × PT = ½ × OT × PR, so 6 × 8 = 10 × PR and PR = 4.8 cm. PQ = 2 × 4.8 = 9.6 cm.
- XY and X′Y′ are two parallel tangents to a circle with centre O, and another tangent AB with point of contact C intersects XY at A and X′Y′ at B. Prove that ∠AOB = 90°. If the radius is 5 cm and AC = 4 cm, find BC. Proof as in Example 11: OA and OB bisect the co-interior angles at A and B, whose sum is 180°, so ∠OAB + ∠OBA = 90° and ∠AOB = 90°. Extension (goes beyond the Circles exercises: it also uses similar triangles from the Triangles chapter). For the length: OC ⊥ AB (Theorem 10.1), and OC is the altitude to the hypotenuse of right triangle AOB. In right triangle OCA, ∠AOC = 90° − ∠OAC = ∠OBC, so right triangles OCA and BCO are similar (AA). Then AC/OC = OC/BC, giving OC2 = AC × BC. So 25 = 4 × BC and BC = 6.25 cm.
Common Mistakes and Examiner Tips
- Adding squares for the tangent length. The right angle is at the point of contact, so the line from the external point to the centre is the hypotenuse. PT2 = OP2 − r2, never OP2 + r2.
- Using the diameter as the radius. A diameter of 10 cm means r = 5 cm. Write r before calculating.
- Stopping at half the chord. In concentric-circle questions, Pythagoras gives half the chord. Double it at the end.
- Calling the two angles complementary. The angle between two tangents and the angle at the centre add up to 180° (supplementary). Show where it comes from: two right angles in quadrilateral OPTQ.
- Skipping the reason for the right angle. Write “∠OPT = 90° (tangent ⊥ radius at the point of contact)” every time. In proofs the reason carries marks.
- Proving Theorem 10.2 with SAS or SSS. The textbook proof uses RHS: right angle (Theorem 10.1), hypotenuse OP common, side OQ = OR. Name the rule and give CPCT for the final step.
- Stating Theorem 10.1 loosely. “Tangent is perpendicular to radius” is incomplete. Say “the tangent at any point of a circle is perpendicular to the radius through the point of contact”.
- Pairing the wrong tangent segments. In a circumscribing quadrilateral or triangle, equal segments share a vertex: AP = AS, not AP = BP. Mark each vertex’s pair with the same letter before adding.
- Giving one tangent from a point inside the circle. Inside: none. On the circle: one. Outside: two.
- Writing a proof without a set-up. Start with “Given”, “To prove” and “Construction”, and draw a neat labelled figure.
Quick Revision Points
- A line and a circle: no common point (non-intersecting), two (secant), one (tangent)
- Tangent: a secant whose two chord end points coincide
- A circle has infinitely many tangents, exactly one at each point
- At most two tangents are parallel to a given secant, one on each side
- Theorem 10.1: tangent at any point ⊥ radius through the point of contact
- Tangents at the two ends of a diameter are parallel
- Tangents through a point: inside 0, on 1, outside 2
- Length of a tangent: segment from the external point to the point of contact
- PT2 = OP2 − r2; OP is always the longest side
- Theorem 10.2: tangents from an external point are equal (RHS congruence)
- The centre lies on the bisector of the angle between the two tangents
- ∠PTQ + ∠POQ = 180° for tangents TP, TQ
- ∠PTQ = 2∠OPQ
- OT is the perpendicular bisector of the chord PQ joining the points of contact
- Concentric circles: chord of larger touching smaller is bisected; length 2√(R2 − r2)
- Quadrilateral around a circle: AB + CD = AD + BC
- Parallelogram around a circle is a rhombus
- Opposite sides of a circumscribing quadrilateral subtend supplementary angles at the centre
- Parallel tangents cut by a third tangent at A and B: ∠AOB = 90°
Weightage in Board Exams
Circles is Chapter 10 of Class 10 Maths. Check the current CBSE course structure for the marks given to it. Because everything rests on two theorems, the question types are predictable.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Tangent length, radius or distance from the centre by Pythagoras; angle between tangents from the angle at the centre; number of tangents from a point; parallel tangents |
| Short answers | Concentric-circle chord; missing side of a circumscribing quadrilateral; equal angles with the chord; simple angle proofs |
| Proofs | Theorem 10.1 and Theorem 10.2 with full reasons; AB + CD = AD + BC; parallelogram is a rhombus; ∠AOB = 90° with parallel tangents; supplementary angles at the centre |
| Case-based | A real setting such as a wheel, a pulley, a park with a gate or a lawn with a fountain, split into parts on tangent lengths and angles |
Practise the two theorem proofs, NCERT Examples 1 to 3 and every question of Exercises 10.1 and 10.2, writing the reason beside each step.
Class 10 Maths · Chapter 10 – swipe through all 10 cards to understand the whole chapter.
A Line and a Circle
A line in the plane of a circle meets it in 0, 2 or exactly 1 point.
No other position of a line with respect to a circle is possible.
- The secant is the whole line; the chord is only the segment between the two points.
- The rope leaving a pulley behaves like a tangent to the pulley.
- A wheel rolls along a line that is a tangent to the wheel.
Tangent as a Special Secant
Turn a secant about a point P: the second meeting point slides to P and the secant becomes the tangent.
The common point is the point of contact; the tangent touches the circle there.
- Only one tangent exists at each point of a circle.
- A circle has infinitely many tangents, one at each point.
- At most two tangents are parallel to a given secant, one on each side.
Tangent ⊥ Radius
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Proof: OP is the shortest distance from O to the tangent, so it is perpendicular.
- Every other point Q of the tangent lies outside the circle, so OQ > OP.
- The line of the radius through the point of contact is called the normal.
- Tangents at the two ends of a diameter are parallel.
Tangents Through a Point
The number of tangents through P depends on where P lies.
Every line through an inside point cuts the circle twice, so none is a tangent.
- Length of a tangent: segment from the external point to the point of contact.
- Two tangents from an outside point touch the circle on either side of OP.
- The tangent line has no length; only the segment PT does.
Length of a Tangent
The radius, the tangent and the line to the centre form a right triangle with the right angle at the point of contact.
OP is the hypotenuse, so it is always the longest side.
- r = 7 cm, OP = 25 cm gives PT = √576 = 24 cm.
- r = 5 cm, OQ = 12 cm gives PQ = √119 cm.
- Halve a given diameter before using the formula.
Equal Tangents
The lengths of tangents drawn from an external point to a circle are equal.
Proof: triangles OQP and ORP are congruent by RHS, then CPCT.
- Pythagoras also works: PQ2 = OP2 − OQ2 = OP2 − OR2 = PR2.
- The centre lies on the bisector of the angle between the tangents.
- OP also bisects the angle QOR at the centre.
Angles Made by Two Tangents
Two right angles in quadrilateral OPTQ link the angle at T with the angle at O.
The two angles are supplementary, never complementary.
- ∠POQ = 110° gives ∠PTQ = 70°.
- Tangents at 80° give ∠POA = 50°.
- OT is the perpendicular bisector of the chord PQ.
Chord Touching the Inner Circle
A chord of the larger circle that touches the smaller circle is bisected at the point of contact.
OP ⊥ AB by Theorem 10.1, and the perpendicular from the centre bisects a chord.
- R = 5 cm, r = 3 cm gives a chord of 8 cm.
- R = 13 cm, r = 5 cm gives a chord of 24 cm.
- Pythagoras gives half the chord: remember to double it.
Circumscribing Figures
Every side is a tangent, so the two tangent segments from each vertex are equal.
A parallelogram circumscribing a circle is a rhombus.
- Opposite sides subtend supplementary angles at the centre.
- Triangle around a circle: tangent segments are s − a, s − b, s − c.
- Radius 4 cm, BD = 8 cm, DC = 6 cm gives AB = 15 cm, AC = 13 cm.
Third Tangent Across Two Parallels
If a tangent at C cuts two parallel tangents at A and B, the segment AB subtends a right angle at the centre.
OA and OB bisect two co-interior angles that add to 180°.
- Half of 180° is 90°, so ∠OAB + ∠OBA = 90°.
- The angle sum of triangle AOB leaves 90° at O.
- This is NCERT Exercise 10.2, Question 9.
📝 Practice Circles - 10 board questions
CBSE previous-year and competency-based · with answers & explanations
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Why B: Step 1 (Theorem 10.1 and Theorem 10.2): ∠OQP = ∠ORP = 90°, ∠QPR = 90°, so ∠QOR = 90°; with OQ = OR = 4 cm and PQ = PR, OQPR is a square of side 4 cm. Step 2 (Pythagoras on the diagonal): OP² = 4² + 4² = 32, so OP = 4√2 cm.
Why not A: 4 cm is the radius OQ, a side of the square, not its diagonal OP.
Why not C: 8 cm is the diameter, which adds two sides instead of using Pythagoras.
Why not D: 2√2 cm is half the diagonal, the distance from the centre of the square to a corner.
Remember: tangents at right angles put the outside point at radius × √2 from the centre.
Why A: Step 1 (Theorem 10.1): ∠OPT = ∠OQT = 90°. Step 2 (angle sum of quadrilateral OPTQ): ∠PTQ = 360° minus 90° minus 90° minus 120° = 60°.
Why not B: 70° does not follow from the angle sum; it is a subtraction slip.
Why not C: 80° does not follow from the angle sum; it is a subtraction slip.
Why not D: 90° is the angle between a radius and its tangent, not the angle between the two tangents.
Remember: angle between the tangents + angle at the centre = 180°.
Why A: by Theorem 10.1 each tangent is perpendicular to the radius at its point of contact. At the two ends of a diameter, both radii lie along the same diameter, so both tangents are perpendicular to the same line and are therefore parallel.
Why not B: each tangent is perpendicular to the diameter, not to the other tangent.
Why not C: a tangent is a line with no fixed length, so equal length has no meaning here.
Why not D: two lines perpendicular to the same line never meet.
Remember: tangents at the ends of a diameter are parallel, and the distance between them is the diameter.
Why B: for tangents PA and PB from P, triangles OAP and OBP are congruent (RHS: OA = OB, OP common, ∠OAP = ∠OBP = 90°), so ∠AOP = ∠BOP and A is true. For a parallelogram ABCD around a circle, equal tangents give AB + CD = AD + BC, so 2AB = 2BC and all sides are equal, so R is true. R is a fact about quadrilaterals and does not explain the equal angles at the centre, so R is not the explanation of A.
Why not A: both statements are true, but R is a different result and does not prove A.
Why not C: R is a standard NCERT result, a parallelogram around a circle must be a rhombus.
Why not D: A follows directly from the congruent right triangles OAP and OBP.
Remember: in A-R first check each statement is true, then ask whether R actually proves A.
Why C: a common tangent touches each circle at one point and does not cut either. When two circles overlap, no line can pass between them without cutting one of them, so the tangents that cross between the circles are impossible. Only the two outer tangents, one on each side, remain.
Why not A: 4 common tangents exist only when the circles are completely separate, with two outer and two crossing tangents.
Why not B: 3 common tangents exist when the circles touch externally, the extra one being the tangent at the touching point.
Why not D: 1 common tangent exists only when the circles touch internally.
Remember: apart 4, touching outside 3, cutting 2, touching inside 1, one inside the other 0.
Why B: Step 1 (Theorem 10.1): ∠OML = 90°, so ∠OMN = 90° minus 70° = 20°. Step 2 (isosceles triangle, OM = ON): ∠ONM = ∠OMN = 20°. Step 3 (angle sum): ∠MON = 180° minus 20° minus 20° = 140°.
Why not A: 120° would need base angles of 30° each, which does not follow from the 70° given.
Why not C: 70° simply copies the angle between the tangent and the chord.
Why not D: 90° is the angle between the radius OM and the tangent, not the angle at the centre.
Remember: the angle a chord subtends at the centre is twice the angle it makes with the tangent at its end.
Why B: Step 1 (Theorem 10.2): tangents from P are equal, so PB = PA = 5 cm. Step 2 (Pythagoras): ∠APB = 90°, so AB is the hypotenuse of right triangle APB and AB² = 5² + 5² = 50, giving AB = 5√2 cm.
Why not A: 5 cm is the tangent PA itself; AB is the hypotenuse and must be longer.
Why not C: 2√5 cm swaps the numbers in 5√2.
Why not D: 10 cm adds PA and PB, which is the path through P, not the straight distance AB.
Remember: perpendicular equal tangents make OAPB a square, and AB is its diagonal, side × √2.
Why A: Step 1 (Theorem 10.1): OA is the radius to the point of contact, so ∠OAT = 90° and OT is the hypotenuse of right triangle OAT. Step 2 (cosine ratio): TA is the side adjacent to the 30° angle at T, so TA = OT × cos 30° = 4 × √3/2 = 2√3 cm.
Why not B: 2 cm = 4 × sin 30° is OA, the radius opposite the 30° angle, not the tangent.
Why not C: 2√2 cm = 4 × 1/√2 uses the 45° value instead of the 30° value.
Why not D: √3 cm is half the correct value, as if √3/2 were multiplied by 2 instead of by 4.
Remember: the tangent is the leg next to the angle at the outside point, so tangent = distance × cos of that angle.
Why A: Step 1 (Theorem 10.1): OB is the radius to the point of contact B, so ∠OBA = 90° and OA is the hypotenuse of right triangle OBA. Step 2 (sine ratio): OB is opposite the 30° angle at A, so OB = OA × sin 30° = 6 × 1/2 = 3 cm.
Why not B: 3√3 cm = 6 × cos 30° is AB, the tangent length, not the radius.
Why not C: 2 cm comes from dividing 6 by 3 instead of by 2.
Why not D: √3 is the value of tan 60°, a trig ratio written down as a length, not a side of this triangle.
Remember: the radius is the leg opposite the angle at the outside point, so radius = distance × sin of that angle.
Why B: A is Theorem 10.1, the tangent at any point is perpendicular to the radius through the point of contact, so A is true. R is Theorem 10.2, the two tangents from an external point are equal, so R is true. But R talks about two tangents from an outside point and says nothing about the angle a tangent makes with a radius, so it cannot explain A. In fact the proof of Theorem 10.2 uses Theorem 10.1, not the other way round.
Why not A: both statements are true, but R is a different fact and does not prove A.
Why not C: R is Theorem 10.2 of the chapter, a true statement.
Why not D: A is Theorem 10.1 of the chapter, a true statement.
Remember: in A-R first check each statement is true, then ask whether R actually proves A.
Chapter Navigation
Previous: Some Applications of Trigonometry Class 10 Notes
Next: Areas Related to Circles Class 10 Notes
Related Chapters in Class 10 Maths
- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
- Triangles Class 10 Notes
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Frequently Asked Questions
A secant is a line that cuts a circle at two points. A tangent is a line that meets the circle at exactly one point, called the point of contact. A tangent is a special case of a secant: when the two end points of the chord cut by the secant come together, the secant becomes the tangent at that point.
It depends on where the point is. From a point inside the circle no tangent can be drawn, because every line through it cuts the circle at two points. Through a point on the circle there is exactly one tangent. From a point outside the circle exactly two tangents can be drawn, and their lengths are equal.
Let XY touch the circle with centre O at P. Take any other point Q on XY. Q lies outside the circle, otherwise XY would be a secant, so OQ is greater than OP. This is true for every point of XY except P, so OP is the shortest distance from O to XY. The shortest segment from a point to a line is perpendicular to it, so OP is perpendicular to XY.
Join the centre O to the point of contact T and to the external point P. The angle at T is 90 degrees because the tangent is perpendicular to the radius. OP is the hypotenuse, so PT² = OP² − r². For example, if OP = 25 cm and r = 7 cm, then PT = √(625 − 49) = √576 = 24 cm.
The RHS rule. Join OP, OQ and OR, where PQ and PR are the tangents. Angles OQP and ORP are right angles by Theorem 10.1, OQ = OR as radii, and OP is the common hypotenuse. So triangles OQP and ORP are congruent by RHS, and PQ = PR by CPCT. The same congruence shows OP bisects the angle between the tangents.
In any quadrilateral drawn around a circle, equal tangent segments from each vertex give AB + CD = AD + BC. In a parallelogram, AB = CD and AD = BC, so the equation becomes 2AB = 2AD, which gives AB = AD. A parallelogram with two adjacent sides equal has all four sides equal, so it is a rhombus.
They are supplementary. If TP and TQ are tangents touching at P and Q, then in quadrilateral OPTQ the angles at P and Q are 90 degrees each. The four angles add to 360 degrees, so angle PTQ + angle POQ = 180 degrees. If the tangents meet at 80 degrees, the angle at the centre is 100 degrees.
