10. Circles

Circles

Chapter 10. Circles

Class 10 Maths Chapter 10 Circles : Multiple Choice Questions , Answer following the Questions , Fill in the blanks , 2 Marks Questions , 3 Marks Question , 4 Marks Question and Solutions :

Class 10 Circles Multiple Choice Questions and Solutions:

SECTION = A

1. In the circle given in figure, the number of tangents parallel to tangent PQ is : [CBSE 2020 basic]

(a) 0 (b) 1 (c) 2 (d) many

Solution: (a) 1

2. In figure, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O . If QPR = 90° , then length of PQ is : [CBSE 2020 standard]

(a) 3 cm (b) 4 cm (c) 2 cm (d) 2 cm

Solution: (b) 4 cm

[ Since OP is the angle bisector of .

In we have ,

So, OPQ is an equilateral triangle .

cm ]

3. In figure, AB is a chord of the circle and AOC is its diameter such that ACB = 50° . If AT is the tangent to the circle at the point A , then BAT is equal to :

(a) 65° (b) 60° (c) 50° (d) 40°

Solution: (c) 50°

[ Since AC perpendicular to AT .

In we have ,

So ,

But ,

3. From an external point P, tangents PA and PB are drawn to a circle with centre O . If , then is :

(a) 30° (b) 35° (c) 45° (d) 25°

Solution: (d) 25° .

[ Since APB is an equilateral triangle .

So,

[ ]

But

]

4. In figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ , then POQ is equal to :

(a) 100° (b) 80° (c) 75° (d) 90°

Solution: (a) 100°

[ Since OP is perpendicular to PR .

POQ is an equilateral triangle .

⇒40°+40°+∠POQ=180°

⇒∠POQ=180°-80°=100° ]

5. In a right triangle ABC , right-angled at B , BC cm and AB cm . The radius of the circle inscribed in the triangle (in cm) is :

(a) 4 (b) 3 (c) 2 (d) 1

Solution:

6. In figure, O is the centre of a circle . PT and PQ are tangents to the circle from an external point P .If , then the measure of is :

(a) (b) (c) (d)

Solution: [ ]

7. A chord of a circle of radius 10 cm subtends a right angle at its centre . The length of the chord is :

(A) cm (B) cm (C) cm (A) cm

Solution:

8. In figure, PQ is a chord of a circle and PT is the tangent at P such that QPT = 60° .Then PRQ is equal to :

(A) 135° (B) 150° (C) 120° (D) 110°

Solution: [ ]

9. In figure, PQ is a tangent at a point C to a circle with centre O . If AB is a diameter and , then is : [ CBSE 2016 ]

(A) (B) (C) (D)

Solution:

10. In figure, PQ is tangent to the circle with centre at O , at the point B .If AOB = 100° , then ABP is equal to : [ CBSE 2020]

(a) 50° (b) 60° (c) 40° (d) 80°

Solution: (c) 40°

[ Since AOB is an equilateral triangle

But ∠ABP+∠OBA=90°

⇒∠ABP=90°-50°

⇒∠ABP=40° ]

Class 10 Circles 3 Marks Questions and Solutions:

SECTION = C

Q1. Prove that the lengths of tangents drawn from an external point to a circle are equal .

Solution: Given O be a centre of the circle and a point P lying outside the circle and two tangents PQ , PR on the circle from P .

To prove :

Construction : We join , and .

Proof : In and we have ,

( Radius of the same circle)

( Common)

[ OAAP and OBBP]

[ R.H.S rule]

(C.P.C.P.) Proved.

Q2. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact .

Solution: Given, a circle with centre O and a tangents XY to the circle at a point P .

To Prove : OP is perpendicular to XY .

Construction : we draw a point Q on XY other than P and join OQ .

Proof : The point Q must lie outside the circle .

Therefore , OQ is longer than the

radius OP of the circle . i.e. , OQ > OP

Every point on the line XY except the point P , OP is the shorter of all the distance of the point O to the points of XY . So, OP is perpendicular to XY .

Class 10 Circles 4 Marks Questions and Solutions:

SECTION = D

Q1. Two tangents TP and TQ are drawn to a circle with centre O from an external point T . Prove that .

Solution: Given, O be a centre of a circle and an external point T and two tangents and to the circle , where P and Q are the points of contact .

To prove :

Proof : Since the length of tangents drawn from an external point to a circle are equal .

i.e. ,

In , we have

and

So , [ OPPT ]

In we have ,

[ From ]

Proved

Q2. Prove that the parallelogram circumscribing a circle is a rhombus .

Solution: Given be a parallelogram and its sides , , and are touch the circle at the points , , and respectively .

To Prove : is a rhombus .

Proof : Since the lengths of tangents drawn from an external point to a circle are equal .

Therefore , (From A )…….

(From B )…….

(From C )…….

(From D )…….

Since be a parallelogram . Thus, and

So ,

Therefore, is a rhombus.

Q3. From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle .Prove that OT is the right bisector of the line segment PQ .

Solution: Given , O be a centre of a circle whose two tangents are TP and TQ .

To Prove : OT is the right bisector of the line segment PQ .

Proof : InOPT and OQT , we have

OP = OQ [ Radius of the circle]

OPT = OPQ = 90° [ OPTP and OQ TQ ]

OT = OT [ common side]

OPT OQT [ R.H.S]

POM = QOM [ C.P.C.T]

InOPM and OQM , we have

OP = OQ [ Radius of the circle]

POM = QOM [ Given ]

OM = OM [ common side]

OPM OQM [ S.A.S]

PM = QM and OMP = OMQ [C.P.C.T ]

OMP + OMQ = 180° [ linear pair of angle ]

OMP + OMP = 180°

OMP = 180° OMP 90° OMP = 90°

OT is the right bisector of the line segment PQ . Proved .

Q4. In figure , a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T ,are of lengths 12 cm and 9 cm respectively .If the area of PQR square cm , then find the lengths of sides PQ and PR .

Solution: