# Free Circles 03 Practice Test - 10th Grade

### Question 1

If AB is the tangent to the circle with center O then, find the measure of ∠OCP.

Given that OP = PC.

30∘

45∘

60∘

15∘

#### SOLUTION

Solution :B

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴∠OPC=90∘

Given, OP = PC.

So, △OPC is an isosceles right angled triangle.⇒∠PCO=∠POC

∠PCO+∠POC+∠OPC=180∘(Angle sum property of a triangle)

∠PCO+∠POC+90∘=180∘

∠PCO+∠POC=90∘

Hence, ∠POC=∠OCP=45∘

### Question 2

In the given figure, LMN is tangent to the circle with centre O. If ∠ PMN = 60∘, find ∠ MOP.

30∘

60∘

90∘

120∘

#### SOLUTION

Solution :D

OMN = 90∘

Radius and tangent are perpendicular at point of contact remaining part of ∠OMN is ∠OMP=90∘−60∘=30∘

OP = OM = Radius

Hence, ∠OMP = ∠OPM = 30∘Therefore, ∠MOP = 120∘.

### Question 3

A circle is inscribed in a triangle with sides 3, 4 and 5 cm. The radius of the circle is

#### SOLUTION

Solution :Join the centre of the circle and the vertices of the triangle. Observe that the sides of the triangle become the tangents to the circle. Hence, the radii of the circle as shown in the question become the heights of the smaller triangles.

According to formula

12 (r) (sum of sides) = area of triangle

12 (r) ( 3+ 4 + 5) = 12 × 3 × 4

r = 1 cm

### Question 4

Two tangents PA and PB are drawn from an external point P to the circle with centre O, such that ∠APB =120∘what is the relation between OP and AP?

OP = 12 AP

AP = 23 OP

OP = 2 AP

OP = AP

#### SOLUTION

Solution :C

Given that ∠APB=120∘

Also, we know that if two tangents are drawn from an external point to a circle, then the line joining the external point and the centre of the circle bisects the angle between the tangents.

⟹∠APO=∠OPB=60∘Thus, cos ∠OPA=cos 60∘=APOP

⟹12 = APOPThus, OP=2AP

### Question 5

State true or false.

Maximum number of tangents to a circle which are parallel to a secant can be two.

True

False

#### SOLUTION

Solution :A

To make tangents parallel to secant only two ways are possible.

### Question 6

Two parallel lines touch a circle at points A and B respectively. The area of the circle is 25π cm2, then the distance between the lines is

5 cm

8 cm

10 cm

25 cm

#### SOLUTION

Solution :C

Given that the area of circle = 25π cm2

⟹πr2=25π, where r is the radius of the circle.

i.e., r2=25⟹r=±5

Since radius is a non-negative quantity, we have r=5 cm

Now, the distance between the two parallel lines

= Diameter of the circle

= 2× Radius of the circle = 10 cm.

### Question 7

State True or False.

The parallelogram circumscribing a circle is a rhombus.

True

False

#### SOLUTION

Solution :A

A parallelogram ABCD circumscribes a circle with centre O.

We know that the lengths of tangents drawn from an exterior point to a circle are equal.

∴ AP = AS, . . . (i) [tangents from A]

BP = BQ, . . . (ii) [tangents from B]

CR = CQ, . . . (iii) [tangents from C]

DR = DS . . . (iv) [tangents from D]

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

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

AB + CD = AD + BC . . . (v)

But we know that in a parallelogram opposite sides are equal.

∴ AD = BC and AB = CD, putting these in (v) we get

2 AB = 2 AD

AB = AD = CD = BC.

Hence, ABCD is a rhombus.

### Question 8

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the major arc.

150∘

30∘

60∘

90∘

#### SOLUTION

Solution :B

Given that AO = AB=OB.

Since all sides are equal, △AOB is equilateral, and hence equiangular. Also, each angle of the triangle equals 60∘.

i.e.,∠AOB = 60∘⟹∠ACB=12AOB

(∵ Angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. )

⟹∠ACB=60∘2=30∘

### Question 9

State true or false.

The incircle of an isosceles triangle ABC, with AB = AC, touches the sides AB, BC, CA at D, E and F respectively. Then E bisects BC.

True

False

#### SOLUTION

Solution :A

We know that the tangents drawn from an external point to a circle are equal.

∴ AD = AF, . . . (i) [tangents from A]

BD = BE, . . . (ii) [tangents from B]

CE = CF . . . (iii) [tangents from C]

Now, AB = AC [given]

⇒ AD + BD = AF + CF

⇒ BD = CF

⇒ BE = CE [using (ii) and (iii)]

⇒ E bisects BC.

### Question 10

The length of the tangent drawn to a circle with radius 7 cm from a point 25 cm away from the centre is =

#### SOLUTION

Solution :∠ABO= 90∘ (point of contact)

Using Pythagoras theorem

AO2= AB2+ BO2

252= AB2+ 72

252

^{ }- 72^{ }= AB224= AB

length of the tangent = 24 cm