CBSE Important Questions Class 10 Maths Chapter 10 – Circles

Class 10 Maths Chapter 10 Circles is part of the CBSE Class 10 Maths syllabus. It explains the idea of a tangent to a circle: a line that touches the circle at exactly one point. Students learn the two main theorems of this chapter, that the tangent at any point of a circle is perpendicular to the radius through the point of contact, and that the lengths of the two tangents drawn from an external point are equal.

These CBSE important questions for Class 10 Maths Chapter 10 cover the tangents drawn from an external point, finding the radius and diameter using the Pythagoras theorem, why the distance from the centre to a tangent equals the radius, a chord of a bigger circle touching a smaller concentric circle, and several assertion-reason questions. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. O is the centre of the circumcircle of triangle PQR. Tangents at P and R intersect at S, and ∠POQ = 130°.
(i) In triangle PSR, ∠PSR = 60°. Which type of triangle is △PSR?
(ii) What is the value of ∠RPQ?

Answer: (i) The lengths of the tangents drawn from an external point to a circle are equal, so PS = RS.
Angles opposite equal sides are equal, so ∠SPR = ∠SRP.
In △SPR: ∠SPR + ∠SRP + 60° = 180°
2∠SPR = 120°, so ∠SPR = 60° = ∠SRP.
All three angles are 60°, so △PSR is an equilateral triangle.
(ii) The radius is perpendicular to the tangent at the point of contact, so ∠OPS = 90°.
∠SPR + ∠OPR = 90°, so ∠OPR = 90° − 60° = 30°.
In △POQ, OP = OQ (radii), so ∠OPQ = ∠PQO = x.
x + 130° + x = 180°, so 2x = 50°, giving x = 25°.
∠RPQ = ∠OPR + ∠OPQ = 30° + 25° = 55°

Q2. The length of the tangent PT drawn from an external point P to a circle is 24 cm. If the distance between P and the centre O is 25 cm, calculate the diameter of the circle. Justify your calculation.

Answer: The tangent to a circle is perpendicular to the radius at the point of contact, so ∠OTP = 90°.
This makes △OTP a right triangle, and we can use the Pythagoras theorem:
OP² = OT² + PT²
25² = r² + 24²
625 = r² + 576
r² = 49
r = 7 cm
Diameter = 2r = 14 cm

Q3. Explain why the distance between the centre of a circle and a tangent to the circle is always equal to the radius.

Answer: The radius drawn to the point of contact is perpendicular to the tangent.
Now, the shortest distance from a point to a line is always measured along the perpendicular from that point to the line.
So the distance from the centre to the tangent is measured along this perpendicular segment, and that segment is exactly the radius.
Therefore the distance between the centre and the tangent is always equal to the radius.

Q4. Read the statements and choose the correct option.
Assertion (A): From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠APB = 60°, then ∠AOB = 120°.
Reason (R): In quadrilateral AOBP, ∠OAP = 90° and ∠OBP = 90°, so ∠APB + ∠AOB = 180°.
(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 A and R are true, and R is the correct explanation of A
In quadrilateral AOBP the sum of all angles is 360°. Since ∠OAP = ∠OBP = 90° (radius perpendicular to tangent):
∠APB + ∠AOB + 90° + 90° = 360°
∠APB + ∠AOB = 180°
So if ∠APB = 60°, then ∠AOB = 120°. The Reason is exactly the working for the Assertion.

Q5. Read the statements and choose the correct option.
Assertion (A): The length of tangent PT from an external point P is 24 cm. If OP = 25 cm, then the diameter of the circle is 14 cm.
Reason (R): A tangent to a circle is perpendicular to the radius through the point of contact.
(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 A and R are true, and R is the correct explanation of A
Because the radius is perpendicular to the tangent, △OTP is right-angled, which lets us apply the Pythagoras theorem: 25² = r² + 24², giving r = 7 and diameter 14 cm. The Reason is the very property that makes the calculation possible.

Q6. Read the statements and choose the correct option.
Assertion (A): The distance between the centre of a circle and any tangent is always equal to the radius.
Reason (R): The shortest distance from a point to a line is the perpendicular distance.
(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 A and R are true, and R is the correct explanation of A
The distance from the centre to the tangent is measured along the perpendicular, and that perpendicular is the radius at the point of contact. So the Reason explains why the Assertion is true.

Q7. In two concentric circles, a chord of the larger circle touches the smaller circle at a point. Prove that the chord is bisected at this point of contact.

Answer: Let the two concentric circles have a common centre O. Let AB be a chord of the larger circle which touches the smaller circle at C.
To prove: AC = CB
Since AB is a tangent to the smaller circle at C, the radius OC is perpendicular to AB, so ∠OCA = 90°.
Now, for the larger circle, AB is a chord and OC is the perpendicular drawn from the centre to this chord.
By the property that the perpendicular from the centre of a circle to a chord bisects the chord, OC bisects AB.
Therefore AC = CB. Hence proved.

Q8. Prove that the lengths of the tangents drawn from an external point to a circle are equal.

Answer: Given: A circle with centre O and an external point P from which two tangents PA and PB are drawn, touching the circle at A and B.
To prove: PA = PB
Construction: Join OA, OB and OP.
Proof: Since the radius is perpendicular to the tangent at the point of contact:
∠OAP = ∠OBP = 90°
In right triangles △OAP and △OBP:
OA = OB (radii of the same circle)
OP = OP (common hypotenuse)
So △OAP ≅ △OBP by the RHS congruence rule.
Therefore PA = PB (corresponding parts of congruent triangles). Hence proved.

Q9. How many tangents can a circle have? Define a secant and a tangent.

Answer: A circle can have infinitely many tangents, because a tangent can be drawn at every point on the circle.
Tangent: A line that touches the circle at exactly one point. That point is called the point of contact.
Secant: A line that cuts the circle at two distinct points.
From a point inside the circle no tangent can be drawn, from a point on the circle exactly one tangent can be drawn, and from a point outside the circle exactly two tangents can be drawn.

Q10. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

Answer: Let the circle touch AB, BC, CD and DA at the points P, Q, R and S respectively.
Using the property that tangents from an external point are equal:
AP = AS (from A)
BP = BQ (from B)
CR = CQ (from C)
DR = DS (from D)
Adding all four:
AP + BP + CR + DR = AS + BQ + CQ + DS
Regrouping:
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC. Hence proved.

Q11. 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 of PQ.

Answer: Since PQ is a tangent at P, the radius OP is perpendicular to PQ, so ∠OPQ = 90°.
In right △OPQ, by the Pythagoras theorem:
OQ² = OP² + PQ²
12² = 5² + PQ²
144 = 25 + PQ²
PQ² = 119
PQ = √119 cm, which is about 10.91 cm

Q12. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer: Given: A circle with centre O and diameter AB. Tangents PQ and RS are drawn at A and B respectively.
To prove: PQ ∥ RS
Proof: Since the radius is perpendicular to the tangent at the point of contact:
OA ⊥ PQ, so ∠OAP = 90°
OB ⊥ RS, so ∠OBS = 90°
Now AB is a straight line (the diameter) cutting the two tangents.
∠OAP and ∠OBS are alternate interior angles, and both are equal to 90°.
Since a pair of alternate interior angles is equal, the two lines must be parallel.
Therefore PQ ∥ RS. Hence proved.

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