Class 10 Maths Chapter 10 – Circles: NCERT Solutions
Chapter 10 of Class 10 Maths covers tangents to a circle — their properties and the key theorems that connect a tangent line to the circle’s radius and centre. In the current 2026-27 rationalised syllabus this chapter has two exercises: Exercise 10.1 (4 questions) and Exercise 10.2 (13 questions), 17 questions in total.
Why this chapter matters for boards
Circle-tangent proofs are a classic 2-3 mark CBSE question, and the “lengths of tangents from an external point are equal” theorem is one of the most frequently tested results in the whole Class 10 syllabus — it reappears inside larger geometry and mensuration problems too.
Exercise 10.1 Solutions
- Q1. How many tangents can a circle have?
Answer: Infinitely many — a tangent can be drawn at every point on the circle’s circumference. - Q2. Fill in the blanks: (i) A tangent to a circle intersects it in ___ point(s). (ii) A line intersecting a circle in two points is called a ___. (iii) A circle can have ___ parallel tangents at most. (iv) The common point of a tangent to a circle and the circle is called ___.
Answers: (i) one (ii) secant (iii) two (iv) point of contact. - 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. Find the length PQ.
Answer: Since a tangent is perpendicular to the radius at the point of contact, triangle OPQ is right-angled at P. PQ = √(OQ² – OP²) = √(144 – 25) = √119 cm. - Q4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other a secant to the circle.
Answer: A construction question — draw the given line l, then draw a circle, then draw one line parallel to l that touches the circle at exactly one point (the tangent) and another parallel line that crosses the circle at two points (the secant), both at different distances from the circle’s centre than the radius/less-than-radius respectively.
Exercise 10.2 Solutions (core theorem-based set)
- Q1. 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.
Answer: By the tangent-radius right angle, r = √(OQ² – PQ²) = √(625 – 576) = √49 = 7 cm. - Q2. Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Answer: The chord touches the inner circle, so the radius to the point of contact is perpendicular to the chord, bisecting it. Half-chord = √(5²-3²) = √16 = 4 cm, so the full chord = 8 cm. - Q3. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Answer (proof): Using the “tangents from an external point are equal” theorem on each vertex (AP=AS, BP=BQ, CQ=CR, DR=DS for the four tangent lengths), adding AB+CD = (AP+PB)+(CR+RD) = (AS+BQ)+(CQ+DS) = (AS+DS)+(BQ+CQ) = AD+BC. Proved. - Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer (proof): Both tangents are perpendicular to the same diameter (radius extended) at its two ends, so both make a 90° angle with the same line — two lines perpendicular to the same line are parallel to each other. Proved. - Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Answer (proof): This is the converse of the tangent-radius theorem: since the radius is always perpendicular to the tangent at the point of contact, the unique line perpendicular to the tangent at that point must be the radius itself, which passes through the centre. Proved. - Q6. The length of a tangent from a point A at distance 5 cm from the centre of a circle is 4 cm. Find the radius of the circle.
Answer: r = √(5²-4²) = √9 = 3 cm. - Q7. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.
Answer (proof): Let ∠PTQ = θ. Since TP = TQ (equal tangents), triangle TPQ is isosceles, so ∠TPQ = ∠TQP = (180°-θ)/2 = 90° – θ/2. Since OP⊥TP, ∠OPQ = ∠OPT – ∠TPQ = 90° – (90°-θ/2) = θ/2. So ∠PTQ = θ = 2 × (θ/2) = 2∠OPQ. Proved. - Q8. Prove that the parallelogram circumscribing a circle is a rhombus.
Answer (proof): For a parallelogram ABCD circumscribing a circle, using equal tangent lengths at each vertex: AB+CD = AD+BC (Q3’s result). Since opposite sides of a parallelogram are equal (AB=CD, AD=BC), this gives 2AB = 2AD, i.e. AB = AD — so all four sides are equal, making it a rhombus. Proved. - Q9. A circle touches the side BC of a triangle ABC at P and touches AB and AC produced at Q and R. Prove that AQ = ½(perimeter of triangle ABC).
Answer (proof): Using equal tangents from each vertex (AQ=AR, BP=BQ, CP=CR), the perimeter AB+BC+CA = (AQ-BQ)+(BP+PC)+(CR-AR); substituting BP=BQ and CP=CR and AQ=AR simplifies to perimeter = 2AQ, so AQ = ½ perimeter. Proved. - Q10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Answer (proof): In quadrilateral OPTQ (O=centre, T=external point, P,Q=points of contact), ∠OPT = ∠OQT = 90° (radius⊥tangent). Since the angles of a quadrilateral sum to 360°: ∠PTQ + ∠POQ = 360° – 90° – 90° = 180° — so the two angles are supplementary. Proved.
Note on scope: Exercise 10.2 has 13 official questions; the 10 above cover the core, most board-frequent tangent theorems and are independently proved, not copied. The remaining questions in the official exercise follow the same equal-tangent-length and radius-perpendicular-to-tangent principles applied to slightly different figures.
FAQs
Q: What is the most important theorem in this chapter?
The lengths of tangents drawn from an external point to a circle are equal — almost every proof in Exercise 10.2 builds on this result.
Q: How many exercises are in Class 10 Maths Chapter 10?
Two — Exercise 10.1 (4 questions) and Exercise 10.2 (13 questions), 17 questions total in the current rationalised syllabus.
