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Circles

Home / Boards / CBSE / Important Questions / Class 10 / Maths / Circles

Class 10 Math Chapter 10
Circles
Important Questions

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Here are some important questions for Class 10 Mathematics Chapter 10, Circles, carefully selected to help students prepare for the CBSE Class 10 Mathematics Examination in 2024-25. These questions cover various types of problems and are designed to assist students in understanding Circles better. By practicing these diverse question types, students can clarify any doubts they may have and improve their problem-solving skills, leading to better performance in the chapter on Circles.

Introduction

In Chapter 10 of Class 10 Mathematics, Circles, In this chapter, we explore the properties of tangents to a circle at the point of contact. Two important properties are:
  • (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact.
  • (Prove) The lengths of tangents drawn from an external point to a circle are equal.

Define a tangent to a circle

A tangent to a circle is a straight line that touches the circle at exactly one point, known as the point of tangency. This point of contact is where the tangent line intersects the circle’s circumference. The tangent line is always perpendicular to the radius drawn to the point of tangency.
cbse class 10 maths circle important questions and answers

Class 10 Circles Important Questions and Answers

Q1. PA and PB are tangents to the circle drawn from an external point P. CD is the third tangent touching the circle at Q. If PB = 12 cm and CQ = 3 cm, then what will be the length of PC?
class 10 circles important questions
Options
(a) 8 cm
(b) 9 cm
(c) 10 cm
(d) 7 cm

Ans.(b) 9 cm
Explanation:
Since, PA and PB are tangents to the circle from point P.
Therefore PA = PB …(i)
Also, CD is the tangent touching the circle at Q.
Therefore, CA = CQ and DQ = DB …(ii)
Since, tangents from external points are equal.
Now, PA = PB = 12 cm
CA = CQ = 3 cm
PC = PA - CA
= 12 - 3 = 9 cm.

Q2. ABC is a right angled triangle, right angled at B such that BC = 6 cm, AB = 8 cm and AC = 10 cm. A circle with centre O is inscribed in ΔABC. The radius of the circle is:
Options
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm

Ans. (b) 2 cm

Explanation:
An incircle is drawn with centre O which touches the sides of the triangle ABC at P, Q and R. OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle.
OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA.
OPBQ is a square. ( ∵ ∠B - 90^o)\\ Let r be the radius of the circle
PB = BQ = r
AR = AP = 8 - r,
CQ = CR = 6 - r
AC = AR + CR
⇒ 10 = 8 - r + 6 - r
⇒ 10 = 14 - 2r
⇒ 2r = 14 - 10 = 4
⇒ r = 2
∴ Radius of the incircle = 2 cm.

circles O2
Q3. The radii of two concentric circles are 13 cm and 8 cm. PQ is the diameter of the bigger circle. QR is a tangent to the smaller circle touching it at R. Find the length PR.

Explanation:
Let the line QR intersect the bigger circle at S.
Join PS.
O is the mid-point of PQ. [∵ PQ is a diameter of the bigger circle] QR is a tangent to the smaller circle and OR is a radius through the point of contact R.

circles_Q3

∴ OR ⊥ QR ⇒ OR ⊥ QS
Since, OR is perpendicular to a chord QS of the bigger circle.
∴ QR = RS
[ ∵ Perpendicular from the centre to a chord bisects the chord]

⇒ R is the mid - point of Qs.
∴ In △QSP, O is the mid-point of PQ and R is the mid-point of QS. \\\text{OR} = \dfrac{1}{2} \text{Ps} \\ [∵ segment joining the mid-points of any two sides of a triangle is half of the third side]

⇒ PS = 2OR = 2 × 8 cm = 16 cm
\\ \text{In right △OQR, OR}^2 + QR^2 = OQ^2\\[4.5 bp] ⇒ 8^2 + QR^2 = 13^2\\[4.5 bp] ⇒ 64 + QR^2 = 169\\[4.5 bp] ⇒ QR^2 = 169 – 64 = 105 \\[4.5 bp] QR = \sqrt{105} \\[4.5 bp] \therefore \space RS = QR = \sqrt{105}\\[4.5 bp] \text{In △PRS, PR}^2 = \text{RS}^2\text{ + PS}^2\\[4.5 bp] = (\sqrt{105})^2 + 16^2\\[4.5 bp] ⇒ PR = 19\space cm.

Q4. In the given figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR = 46°, then find ∠QOR.
Circles Q4

Explanation:
Join OP, such that it bisects ∠QPR.
Thus, ∠OPQ = ∠OPR = 23°
Also, ∠OQP = ∠ORP = 90°
So, ∠QOP = ∠ROP
= 180° – (90° + 23°)
= 180° – 113° = 67°
Hence, ∠QOR = ∠QOP + ∠ROP
= 67° + 67°
= 134°

Q5. From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of △PCD.

Explanation:
Given, PA and PB are the tangents drawn from a point P outside the circle with centre O.
CD is another tangent to the circle at point E which intersects PA and PB at C and D respectively.

Circles Q5

Here, PA = 14 cm
PA and PB are the tangents to the circle from P
So, PA = PB = 14 cm
Now, CA and CE are the tangents from C to the circle.
∴ CA = CE …(i)
Similarly , DB and DE are the tangents from D to the circle.
∴ DB = DE …(ii)
Now, perimeter of △PCD
= PC + PD + CD
= PC + PD + CE + DE
= PC + CE + PD + DE
= PC + CA + PD + DB
{From (i) and (ii)}
= PA + PB
= 14 + 14
= 28 cm

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CBSE Class 10 Maths Chapter wise Important Questions

making of global world class 10 important questions

Conclusion

If you want to get better at understanding the Circles chapter, you should check out oswal.io. This website offers lots of practice questions made to help you learn more effectively. By practicing these questions, you can improve how well you understand circles and get better at solving problems related to them. It’s a really helpful way to practice and feel more confident when dealing with circle concepts.

Frequently Asked Questions

Ans: The diameter of a circle is twice the length of its radius. In other words, diameter = 2 × radius.
Ans: The circumference of a circle (C) is given by the formula C = 2πr, where ‘r’ is the radius. Also, the diameter (d) is twice the radius, so d = 2r.
Ans: Angle between a tangent and a radius drawn to the point of tangency is 90 degrees, making them perpendicular to each other.
Ans: In a circle, tangents drawn from an external point are equal in length. They are both perpendicular to the radius drawn to the point of contact on the circle.
Ans: The angle sum of a quadrilateral formed by four tangents drawn to a circle is 360 degrees.
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