# Free Circles 01 Practice Test - 10th Grade

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 ___ cm.

#### 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=62549=576

QP=24 cm
The length of the tangent is 24 cm.

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

60

B.

70

C.

80

D.

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=3609090110

PTQ=70

PTQ=70

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.

A.

True

B.

False

#### SOLUTION

Solution : A Given that AO = 10 cm, BO = 6 cm

Applying Pythagoras theorem

AB2=AO2BO2

AB2=10262

AB2=10036=64

AB=8 cm

Through any given set of three distinct points A, B, C it is possible to draw at most ___circle(s).

#### 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). .

State true or false.
Maximum number of tangents parallel to a given secant of a circle are four.

A.

True

B.

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.

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.

A. 10 cm
B. 5 cm
C. 7 cm
D. 12 cm

#### 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=132122
OT2=25
OT=25
OT=5
So, radius of the circle is 5 cm

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. A.

8 cm

B.

16 cm

C.

10 cm

D.

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=OP2OB2=10262cm=64cm

Thus, the length of BP
=64cm = 8 cm.

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? A.

BD = 41.42 cm

B.

OD = 141.42 cm

C.

PD = 100 cm

D.

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=OPPD

Tan45=100PD (OP=1m=100cm)

1=100PD

PD=100 cm
Similarly we can show that PC = 100 cm

Now, sinODP=OPOD

sin45=0.7071=100OD

i.e., OD=1000.7071

OD=141.42 cm

But, BD=ODOB

BD=141.42100=41.42 cm

AC+CP=BD+100 (since AC = BD and PC= 100 cm)
=41.42+100=141.42cm

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.

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

55

B.

110

C.

120

D.

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