Circles Test

Result

Circles Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Two equal circles in the same plane cannot have the following number of common tangents.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

E
None of these

Solution

We observe that two equal circles in a plane cannot have only one common tangent.
Question 2
1 / -0

Which circle is an example of point of contact of a circle?

A

B

C

D

Solution

The figure 3 is an example of point of contact of the circle.
Since the point of contact in a circle is a point where the tangent touches the circle at one point.
Question 3
1 / -0

The point of contact of the circle is

A
P

B
C

C
Q

D
A

Solution

The point of contact of the circle and tangent is $$A$$.
The point of contact in a circle is a point where the tangent touches the circle.
Question 4
1 / -0

Consider the following statements and identify which are correct:
i) A secant to a circle can act as a chord.
ii) A chord cannot be a secant to the circle.

A
i only

B
ii only

C
both are correct

D
both are wrong

Solution

A chord of a circle is a straight line segment whose endpoints both lie on the circle
where the secant is a line stretching to infinity in both directions.
A chord is a line segment that only covers the part inside the circle.
therefore A secant to a circle can act as a chord.
and chord cannot be a secant to the circle.
option C is the answer.
Question 5
1 / -0

How many point of contacts are there in a circle?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

The point of contact in a circle is a point where the tangent touching the circle at one point.
There are 3 point of contacts are in the circle.
$$D, E$$ and $$F$$ are the three point of contact in a circle.
Question 6
1 / -0

What is a line passing through two points on a circle called?

A
Secant

B
Digonal

C
Radius

D
Tangent

Solution

A line passing through two points on a circle called secant.
Here $$AB$$ is a secant.
Question 7
1 / -0

A circle and a tangent to the circle have in common

A
one point

B
two points

C
no point

D
many points

Solution

A circle and a tangent to the circle have in common only one point.
Question 8
1 / -0

A line which intersects a circle in only one point is called

A
A Secant

B
A tangent

C
A chord

D
A diameter

Solution

A line which intersects a circle in only one point is called A tangent
Question 9
1 / -0

To draw a pair of tangents to a circle which are inclined to each other at an angle of $$60^0$$, it is required to draw tangents at endpoints of those two radii of the circle, the angle between them should be

A
$$135^{0}$$

B
$$90^{0}$$

C
$$60^{0}$$

D
$$120^{0}$$

Solution

Given:
$$ PA$$ & $$PB$$ are two tangents drawn from P to a circle with centre O at A & B respectively.
$$\angle APB={ 60 }^{ o }$$
To find out-
$$ \angle AOB$$
Solution-
$$ PA$$ & $$PB$$ are two tangents drawn from P to the circle at A & B, respectively.
$$ \therefore PA=PB\Longrightarrow \Delta PAB$$ is isosceles.
i.e $$\angle PAB=\angle PBA.$$
or $$\angle PAB+\angle PBA=2\angle PAB$$ .....(i)
Now, $$\angle PAB+\angle PBA+\angle APB={ 180 }^{ o }$$ ....(angle sum property of triangles)
$$ \Longrightarrow \angle PAB+\angle PBA{ +60 }^{ o }={ 180 }^{ O }$$
$$\Longrightarrow 2\angle PAB={ 120 }^{ o }$$ ...(from i)
$$ \therefore \angle PAB={ 60 }^{ o }=\angle PBA$$ .........(ii)
Again, $$\angle OAP={ 90 }^{ o }$$ ....(angle between a radius and the tangent at the point of contact.)
$$ \therefore \angle OAB=\angle OAP-\angle PAB={ 90 }^{ o }-{ 60 }^{ o }$$ .... (from ii) .........(iii)
Since, $$OA=OB$$ ...(radii of the same circle)
$$ \therefore \Delta OAB$$ is isosceles
$$\Longrightarrow \angle OAB=\angle OBA={ 30 }^{ o }$$ ....(from iii)
So, $$\angle AOB={ 180 }^{ o }-(\angle OAB+\angle OBA)={ 180 }^{ o }-{ 30 }^{ o }-{ 30 }^{ O }={ 120 }^{ O }$$ ...(angle sum property of triangles)
Question 10
1 / -0

In the figure, AB is a chord of the circle and AOC is its diameter such that $$\angle ACB =50^{0}$$. If AT is the tangent to the circle at the point A, then $$\angle BAT$$ is equal to

A
$$65^{0}$$

B
$$60^{0}$$

C
$$50^{0}$$

D
$$40^{0}$$

Solution

The line AC passes through O. So it is a diameter of the circle.

$$ \therefore \angle ABC$$ is an angle inscribed in a semicircle.

$$ \therefore \angle ABC={ 90 }^{ O }.$$

Now $$\angle ABC+\angle ACB+\angle BAC={ 180 }^{ O }$$ ...(sum of the { angles of a $$\Delta ={ 180 }^{ O })$$

$$\Rightarrow { 50 }^{ O }+{ 90 }^{ O }+\angle BAC={ 180 }^{ O }$$

$$\Rightarrow \angle BAC={ 40 }^{ O }$$

Again, $$AT$$ is a tangent to the circle at $$A.$$
$$ \therefore \angle CAT={ 90 }^{ O }$$

$$\Rightarrow \angle BAT+\angle CAB={ 90 }^{ O }$$

$$\Rightarrow \angle BAT+{ 40 }^{ o }={ 90 }^{ O }$$

$$\Rightarrow \angle BAT={ 50 }^{ O }.$$