# Free Circles 01 Practice Test - 10th Grade

### Question 1

A point P is 25 cm from the centre of a circle. The radius of the circle is 7 cm and length of the tangent drawn from P to the circle is x cm. The value of x is

#### SOLUTION

Solution :Given that OP = 25 cm and OQ = 7 cm.

To find the length of PQ, apply Pythagoras theorem in △OPQ since ∠OQP=90∘

OQ2+QP2=OP2

72+QP2=252

QP2=625–49=576

QP=24 cm

The length of the tangent is 24 cm.

### Question 2

If TP and TQ are two tangents to a circle with center O such that ∠POQ=110∘, then, ∠PTQ is equal to:

60∘

70∘

80∘

90∘

#### SOLUTION

Solution :B

Given that ∠POQ=110∘

Note that ∠OQT=∠OPT=90∘

(∵ A tangent at any point of a circle is perpendicular to the radius at the point of contact)Also, ∠TQO+∠QOP+∠OPT+∠PTQ=360∘(∵ Sum of interior angles of a quadrilateral is 360 degrees)

⟹∠PTQ=360∘–90∘–90∘–110∘

⟹∠PTQ=70∘

∴∠PTQ=70∘

### Question 3

State true or false.

The length of the tangent drawn from a point 10 cm away from the center of a circle of radius 6 cm is 8 cm.

True

False

#### SOLUTION

Solution :A

Given that AO = 10 cm, BO = 6 cm

Applying Pythagoras theorem

AB2=AO2–BO2

AB2=102–62

⇒AB2=100–36=64

AB=8 cm

### Question 4

Through any given set of three distinct points A, B, C it is possible to draw at most

#### SOLUTION

Solution :At most one circle can be drawn through a given set of three distinct points. These three points will then be referred to as 'concyclic points' (Lying on the same circle). .

### Question 5

State true or false.

Maximum number of tangents parallel to a given secant of a circle are four.

True

False

#### SOLUTION

Solution :B

Take any secant like AB as shown, there cannot be more than 2 tangents parallel to it. So, the answer is 2 not 4.

### Question 6

A point P is 13 cm from the centre of the circle. The length of the tangent drawn from P to the circle is 12cm. Find the radius of the circle.

#### SOLUTION

Solution :B

Since,

tangent to a circle is perpendicular to the radius through the point of contact

So, ∠OTP=900

So, in triangle OTP

(OP)2=(OT)2+(PT)2

132=(OT)2+122

(OT)2=132−122

OT2=25

OT=√25

OT=5

So, radius of the circle is 5 cm

### Question 7

Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radii 8 cm and 6 cm respectively, as shown in the figure. If AP = 6 cm then find the length of BP.

8 cm

16 cm

10 cm

6 cm

#### SOLUTION

Solution :A

We have

OA ⊥ AP and OB ⊥ BP [ The tangent at any point of a circle is perpendicular to the radius through the point of contact].

Join OP.

In right Δ OAP, we have

OA = 8 cm, AP = 6 cm

∴ OP2=OA2+AP2 [by Pythagoras theorem]

⇒ OP=√OA2+AP2=√82+62cm=√100cm=10 cm

In right Δ OBP, we have

OB = 6 cm, OP = 10 cm

∴ OP2=OB2+BP2

[by Pythagoras' theorem]⇒ BP=√OP2−OB2=√102−62cm=√64cm

Thus, the length of BP

=√64cm = 8 cm.

### Question 8

In a circle, O is the centre and ∠COD is right angle. AC = BD and CD is the tangent at P. Which of the following are true, if the radius of the circle is 1 m?

BD = 41.42 cm

OD = 141.42 cm

PD = 100 cm

AC + CP = 141.42 cm

#### SOLUTION

Solution :A, B, C, and D

Given that AC = BD.

OA = OB (radius)Adding the above equations,

AC + OA = BD + OB

⟹OC = OD

⟹∠ODC=∠OCD (Property of isosceles triangles)

Given ∠COD=90∘

Therefore, ∠ODC=∠OCD=45∘ (Base angles are equal)

Join OP.

Since a tangent at any point of a circle is perpendicular to the radius at the point of contact, we have OP ⊥ CD.

Consider right angled triangle ODP,

tan ∠ODP=OPPDTan45∘=100PD (∵OP=1m=100cm)

⟹1=100PD

⇒PD=100 cm

Similarly we can show that PC = 100 cmNow, sin∠ODP=OPOD

⟹sin45∘=0.7071=100OD

i.e., OD=1000.7071

⇒OD=141.42 cm

But, BD=OD−OB

⇒BD=141.42–100=41.42 cm

AC+CP=BD+100 (since AC = BD and PC= 100 cm)

=41.42+100=141.42cm

### Question 9

A line that touches a circle at only one point is called

#### SOLUTION

Solution :A line that touches a circle at only one point is called a tangent.

### Question 10

In the adjoining figure 'O' is the center of circle, ∠CAO = 25∘ and ∠CBO = 35∘. What is the value of ∠AOB?

55∘

110∘

120∘

Data insufficient

#### SOLUTION

Solution :C

In ΔAOC,

OA=OC --------(radii of the same circle)

∴ΔAOC is an isosceles triangle

→∠OAC=∠OCA=25∘----- (base angles of an isosceles triangle )

In ΔBOC,

OB=OC --------(radii of the same circle)

∴ΔBOC is an isosceles triangle

→∠OBC=∠OCB=35∘ -----(base angles of an isosceles triangle )

∠ACB=25∘+35∘=60∘

∠AOB=2×∠ACB ----(angle at the center is twice the angle at the circumference)

= 2×60∘

=120∘