CBSE X Maths — Circles: Tangents: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit IV: Geometry

Circles — Tangents

A tangent touches a circle at exactly one point. Two simple facts about tangents — that they’re perpendicular to the radius, and that two tangents from the same external point are equal in length — unlock most problems in this topic.

1. Tangent Perpendicular to the Radius

Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

This means that if PA is a tangent to a circle with centre O, touching at A, then OA⊥PA — so triangle OAP is right-angled at A, and Pythagoras’ theorem applies directly.

Worked example — using the right angle at the point of contact A tangent PA is drawn to a circle of radius 5 cm from an external point P, with PA=12 cm. Find OP.
Since OA⊥PA, triangle OAP is right-angled at A: OP² = OA² + PA² = 5² + 12² = 25+144 = 169
OP = 13 cm

2. Equal Tangents from an External Point

Theorem: The lengths of the two tangents drawn from an external point to a circle are equal.

Worked example — finding the radius From a point 25 cm from the centre of a circle, the length of the tangent is 24 cm. Find the radius.
r² = OP² − PT² = 25² − 24² = 625 − 576 = 49
r = 7 cm

Worked example — angle between tangents and radii Two tangents from an external point P are inclined to each other at 60°. Find the angle between the two radii drawn to the points of contact.
In quadrilateral OAPB (O=centre, A,B=points of contact), the angles at A and B are each 90° (tangent⊥radius). The angles of a quadrilateral sum to 360°:
90 + 90 + 60 + ∠AOB = 360 → ∠AOB = 120°

Worked example — tangent lengths in a circumscribed triangle A triangle ABC with AB=10, BC=8, CA=6 circumscribes a circle. Find the tangent length from each vertex.
Semi-perimeter s = (10+8+6)/2 = 12. Using the standard result (tangent length from a vertex = s − opposite side):
From A: s−BC = 12−8 = 4. From B: s−CA = 12−6 = 6. From C: s−AB = 12−10 = 2.
Check: AB = 4+6=10 ✓, BC = 6+2=8 ✓, CA = 2+4=6 ✓

Common mistakes to watch for

Forgetting that the right angle in a tangent problem is always at the point of contact, not at the external point.
In OAPB-type quadrilateral problems, forgetting that both angles at the points of contact are 90° — this is what makes the angle sum approach work.
Not simplifying Pythagoras’ theorem correctly when solving for the radius vs. the tangent length vs. the distance to the centre — always identify which one is the hypotenuse (it’s always the line from the external point to the centre).

Key Vocabulary Recap

Tangent — a line touching a circle at exactly one point

Point of contact — the single point where a tangent touches the circle

Syllabus Reference

This note covers CBSE Class X Mathematics Unit IV, Circles: the tangent to a circle at the point of contact, proof that it is perpendicular to the radius through that point, and proof that the lengths of tangents from an external point are equal.

Practice questions

CBSE Class X Mathematics · Unit IV: Circles — Practice Question Bank (no answers)

Circles — Practice

Covers: tangent-radius perpendicularity, and equal tangents from an external point. 10 questions.

A tangent PA is drawn to a circle of radius 5 cm from an external point P such that PA=12 cm. Find OP (O is the centre).

Prove that the two tangents drawn from an external point to a circle are equal in length.

From a point 25 cm from the centre of a circle, the length of the tangent to the circle is 24 cm. Find the radius.

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

A circle of radius 7 cm has a tangent PT drawn from an external point P, where PT=24 cm. Find the distance OP.

Two tangents to a circle from an external point are inclined to each other at 60°. Find the angle between the two radii drawn to the points of contact.

A triangle ABC with AB=10 cm, BC=8 cm, CA=6 cm circumscribes a circle. Find the length of the tangent from each vertex to the circle.

Prove that if a circle touches all four sides of a quadrilateral, then the sums of the two pairs of opposite sides are equal.

PA and PB are tangents from an external point P to a circle with centre O, such that ∠APB=80°. Find ∠AOB.

Q10

A point P is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from P to the circle.

More Practice

Q11

In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm, find the radius of the smaller circle.

Q12

PQ is a tangent from an external point P to a circle with centre O and radius 6 cm, such that OP=10 cm. Find the length of PQ.

Q13

Two circles touch externally. The sum of their areas is 130π cm² and the distance between their centres is 14 cm. Find the radii.

Q14

If a hexagon ABCDEF circumscribes a circle, prove that AB+CD+EF = BC+DE+FA.

Q15

Prove that a parallelogram circumscribing a circle is a rhombus.

Q16

A circle touches all four sides of a quadrilateral ABCD. If AB=6 cm, BC=7 cm, CD=4 cm, find AD.

Q17

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

Q18

From an external point P, the tangent length to a circle of radius 6 cm is 8 cm. Find the distance OP.

Q19

PA and PB are tangents from an external point P to a circle of radius 5 cm, with ∠AOB=120°. Find the length PA.

Q20

A triangle ABC circumscribes a circle of radius 3 cm, and the tangent point D on BC divides it into BD=6 cm and DC=9 cm. If the area of triangle ABC is 63 cm², find the sides AB and AC.

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit IV: Circles — Solutions

Circles — Practice: Solutions

Radius 5, PA=12. Find OP.

OA⊥PA, so use Pythagoras

OP²=5²+12²=169

✓OP = 13 cm

Prove tangents from an external point are equal.

✓Let PA and PB be tangents from external point P to a circle with centre O, touching at A and B. In triangles OAP and OBP: OA=OB (radii), OP is common, and ∠OAP=∠OBP=90° (tangent⊥radius). By RHS congruence, ΔOAP≅ΔOBP, so PA=PB.

Distance 25, tangent 24. Find radius.

r²=OP²−PT²

25²−24²=625−576

✓r = 7 cm

Prove tangent⊥radius at point of contact.

✓Let PA be a tangent to a circle with centre O, touching at A. Take any other point B on the tangent line. Since B lies outside the circle (the tangent meets the circle only at A), OB > OA. This means OA is the shortest distance from O to the tangent line, and the shortest distance from a point to a line is always the perpendicular distance — so OA⊥PA.

Radius 7, PT=24. Find OP.

OP²=OT²+PT²

7²+24²=49+576=625

✓OP = 25 cm

Tangents inclined at 60°. Find ∠AOB.

Angles of quadrilateral OAPB sum to 360°, with 90° at A and B

90+90+60+∠AOB=360

✓∠AOB = 120°

AB=10,BC=8,CA=6, circumscribed circle. Find tangent lengths.

Semi-perimeter s=(10+8+6)/2=12. Tangent length from a vertex = s − opposite side

From A: 12−8=4; From B: 12−6=6; From C: 12−10=2

✓4 cm from A, 6 cm from B, 2 cm from C (check: 4+6=10, 6+2=8, 2+4=6 ✓)

Prove opposite sides of a tangential quadrilateral sum equally.

✓Let a circle touch sides AB, BC, CD, DA of quadrilateral ABCD at P, Q, R, S. Using equal tangents from each vertex: AP=AS, BP=BQ, CQ=CR, DR=DS. Then AB+CD = (AP+PB)+(CR+RD) = (AS+BQ)+(CQ+DS) = (AS+DS)+(BQ+CQ) = AD+BC. So AB+CD = AD+BC — the sums of opposite sides are equal.

∠APB=80°. Find ∠AOB.

Same quadrilateral angle-sum approach as Q6

90+90+80+∠AOB=360

✓∠AOB = 100°

Q10

Distance 13, radius 5. Find tangent length.

PT²=OP²−OT²

13²−5²=169−25=144

✓PT = 12 cm

More Practice — Solutions

Q11

Chord 8cm of larger circle(r=5) touches smaller circle.

Half-chord=4cm; the radius to the point of tangency is ⊥ to the chord

r²=5²−4²=9

✓r = 3 cm

Q12

r=6, OP=10. Find PQ.

PQ²=OP²−OQ²

100−36=64

✓PQ = 8 cm

Q13

Circles touch externally, sum of areas 130π, distance 14. Find radii.

r₁+r₂=14 (touching externally); r₁²+r₂²=130

(r₁+r₂)²=196=130+2r₁r₂ → r₁r₂=33

Solve t²−14t+33=0. Discriminant=196−132=64

t=(14±8)/2

✓Radii are 11 cm and 3 cm

Q14

Prove hexagon tangent-sum property.

✓Let the circle touch the six sides at points such that equal tangents from each vertex are labelled appropriately. Using equal tangents from each of the 6 vertices and summing alternating sides (exactly as in the quadrilateral case), AB+CD+EF and BC+DE+FA both equal the sum of all twelve tangent segments’ halves taken once each — so AB+CD+EF = BC+DE+FA.

Q15

Prove parallelogram circumscribing a circle is a rhombus.

✓For any tangential quadrilateral, AB+CD=AD+BC. Since ABCD is a parallelogram, AB=CD and AD=BC. Substituting: 2AB=2AD → AB=AD. Since opposite sides of a parallelogram are also equal, all four sides are equal — it is a rhombus.

Q16

AB=6,BC=7,CD=4. Find AD.

AB+CD=AD+BC

6+4=AD+7

✓AD = 3 cm

Q17

Prove ∠PTQ=2∠OPQ.

✓Since OP=OQ (radii), ΔOPQ is isosceles, so ∠OPQ=∠OQP. In ΔOPT (right-angled at P), ∠OPT=90°, so ∠OPQ+∠QPT=90°. Also, since ΔOPT≅ΔOQT (RHS), PT bisects ∠PTQ, and OT bisects ∠POQ. Using the angle sum in quadrilateral OPTQ: 90+90+∠PTQ+∠POQ=360, so ∠PTQ+∠POQ=180. Also in isosceles ΔOPQ, ∠POQ=180−2∠OPQ. Substituting: ∠PTQ+(180−2∠OPQ)=180 → ∠PTQ=2∠OPQ.

Q18

r=6, tangent=8. Find OP.

OP²=r²+PT²

36+64=100

✓OP = 10 cm

Q19

r=5, ∠AOB=120°. Find PA.

In quadrilateral OAPB, ∠APB=360−90−90−120=60°. By symmetry, ∠OPA=30°

tan30°=OA/PA → PA=OA/tan30°=5/(1/√3)

✓PA = 5√3 cm

Q20

r=3, BD=6, DC=9, area=63. Find AB, AC.

s=Area/r=63/3=21. BC=BD+DC=15

Let tangent length from A=x. AB=x+6, AC=x+9

Perimeter=2s=42: (x+6)+15+(x+9)=42

2x+30=42 → x=6

✓AB = 12 cm, AC = 15 cm

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