Important Questions Class 10 Maths Chapter 10 Circles

A circle in mathematics can be defined as a geometric figure where all the points in a plane ( a flat surface that can infinitely go in both directions) are equidistant from the centre (fixed spot). Look at a bicycle wheel where all the spokes are equidistant from the centre of the wheel. The term circle is obtained from Greek syllables. In mathematics, the exploration of the circle has led to many developments in the field of geometry, astronomy, calculus, etc. In day-to-day life, the circle is the foundation for the wheel which caused the invention of many modern pieces of machinery and a tangent is a line on a plane that intersects the circle at only one point. The term tangent is also obtained from Greek syllables which means “to touch”.

The important question for class 10 maths chapter 10 Circles can be obtained from the following topics:

• Defining tangent to a circle 
• Defining secant
• Theorem of tangents

Important Questions Chapter 10 – Circles

Q1. In the given figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR = 46°, then calculate ∠QOR

Solution:

∠OQP = 900

∠ORP = 90°

∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° …[Angle sum property of a quad.

90° + 46° + 90° + ∠QOR = 360°

∠QOR = 360° – 90° – 46° – 90° = 134°

Q2. In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB

Solution:

PA = PB …[∵ Tangents drawn from external point are equal

∠OAP = ∠OBP = 90°

∠OAB = ∠OBA … [Angles opposite equal sides

∠OAP + ∠AOB + ∠OBP + ∠APB = 360° … [Quadratic rule

90° + ∠AOB + 90° + 50° = 360°

∠AOB = 360° – 230°

= 130°

∠AOB + ∠OAB + ∠OBA = 180° … [∆ rule

130° + 2∠OAB = 180° … [From (i)

2∠OAB = 50°

⇒ ∠OAB = 25°

Q3. From an external point P, tangents PA and PB are drawn to a circle with centre 0. If ∠PAB = 50°, then find ∠AOB

Solution:

PA = PB …[∵ Tangents drawn from external point are equal

∠PBA = ∠PAB = 50° …[Angles equal to opposite sides

In ∆ABP, ∠PBA + ∠PAB + ∠APB = 180° …[Angle-sum-property of a ∆

50° + 50° + ∠APB = 180°

∠APB = 180° – 50° – 50° = 80°

In cyclic quadrilateral OAPB

∠AOB + ∠APB = 180° ……[Sum of opposite angles of a cyclic (quadrilateral is 180°

∠AOB + 80o = 180°

∠AOB = 180° – 80° = 100°

Q4. In the given figure, the sides AB, BC and CA of a triangle ABC touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC (in cm)

Solution:

AP = AR = 4 cm

RC = 11 – 4 = 7 cm

RC = QC = 7 cm

BQ = BP = 3 cm

BC = BQ + QC

= 3 + 7 = 10 cm

Q5. Find the perimeter (in cm) of a square circum scribing a circle of radius a cm. 

Solution:

Radius = R

AB = a + a = 2a

∴ Perimeter = 4(AB)

= 4(2a)

= 8a cm

Q6. In the figure, AB is the diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ. 

Solution:

∠ABQ = ½  ∠AOQ =58°/2 = 29°

∠BAT = 90° ….[Tangent is ⊥ to the radius through the point of contact

∠ATQ = 180° – (∠ABQ + ∠BAT)

= 180 – (29 + 90) = 180° – 119° = 61°

Q7. In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD

Solution:

AB + CD = AD + BC

6 + 8 = AD + 9

14 – 9 = AD ⇒ AD = 5 cm

Q8. In figure, a quadrilateral ABCD is drawn to circum- DA scribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, RA and S respectively. Prove that: AB + CD = BC + DA

Solution:

AP = AS ……(i) (Tangents drawn from an external point are equal in length

BP = BO …(ii)

CR = CQ ….(iii)

DR = DS ..(iv)

By adding (i) to (iv)

(AP + BP) + (CR + DR) = AS + BQ + CQ + DS

AB + CD = (BQ + CQ) + (AS + DS)

∴ AB + CD = BC + AD (Hence proved)

Q9. The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC

Solution:

Given: The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F respectively.

AB = AC

To prove: BD = CD

Proof: AF = AE ..(i)

BF = BD …(ii)

CD = CE …(iii)

Adding (i), (ii) and (iii), we get

AF + BF + CD = AE + BD + CE

⇒ AB + CD = AC + BD

But AB = AC …[Given]

∴ CD = BD

Q10. In Figure, common tangents AB and CD to the two circles with, centres O1 and O2 intersect at E. Prove that AB = CD

Solution:

EA = EC …(i) ….[Tangents drawn from an external point are equal

EB = ED …(ii)

EA + EB = EC + ED …[Adding (i) & (ii)

∴ AB = CD (Hence proved)