# CBSE Class 10-Mathematics: Questions and Answers Chapter – 10 Circles Part 6 (For CBSE, ICSE, IAS, NET, NRA 2023)

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**Question 6**:

If tangents PA and PB from a point P to a circle with center are inclined to each other at angle of , then is equal to:

(A)

(B)

(C)

(D)

**Answer**:

(A)

[The tangent at any point of a circle is to the radius through the point of contact]

[Centre lies on the bisector of the angle between the two tangents]

In ,

[Angle sum property of a triangle]

**Question 7**:

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

**Answer**:

Given: PQ is a diameter of a circle with center O.

The lines AB and CD are the tangents at P and Q, respectively.

To Prove:

Proof: Since AB is a tangent to the circle at P and OP is the radius through the point of contact.

[The tangent at any point of a circle is to the radius through the point of contact]

CD is a tangent to the circle at Q and OQ is the radius through the point of contact.

[The tangent at any point of a circle is to the radius through the point of contact]

From eq. (i) and (ii) ,

But these form a pair of equal alternate angles also,

**Question 8**:

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.

**Answer**:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact and the radius essentially passes through the center of the circle, therefore the perpendicular at the point of contact to the tangent to a circle passes through the center.

**Question 9**:

The length of a tangent from a point A at distance 5 cm from the center of the circle is . Find the radius of the circle.

**Answer**:

We know that the tangent at any point of a circle is to the radius through the point of contact.

[By Pythagoras theorem]