Circles Test - 13

Here \(A B\) is a diameter of the circle with centre \(O\), two tangents \(P Q\) and \(R S\) drawn at points \(A\) and \(B\) respectively:

Radius will be perpendicular to these tangents. 

Thus, \(O A \perp R S\) and \(O B \perp P Q\) 

\(\angle {OAR}=\angle {OBP}=\angle{OBQ}=90^{\circ}\)

Therefore, \(\angle {OAR}=\angle {OBQ}\) (Alternate interior angles) 

\(\angle {OAS}=\angle {OBP}\) (Alternate interior angles)

Since alternate interior angles are equal, lines \({PQ}\) and \({RS}\) will be parallel.

Hence, the correct option is (B).

Distance between the two parallel tangent to a circle \(=\) diameter \(={PQ}={PO}+\mathrm{OQ}\)

\(=2 \times \)radius

\(=2 \times 4=8\) cm

Hence, the correct option is (C).

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

Draw a circle with centre \({O} ; {PA}\) and \({PB}\) are two tangents to the circle drawn from an external point \(P\). Join \(O A, O B,\) and \(O P\).

\(OA\perp PA\)

\({OB} \perp {PB}\)

In \(\triangle{OPA}\) and \(\triangle {OPB}\)

\(\angle {OPA}=\angle {OPB}\) (Using (1))

\({OA}=\mathrm{OB}\) (Radii of the same circle) 

\({OP}={OP}\) (Common side) 

Therefore \(\triangle {OPA} \cong \triangle {OPB}\) (RHS congruency criterion) 

\({PA}={PB}\)

(Corresponding parts of congruent triangles are equal) Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.

The length of tangents drawn from any external point are equal.

Hence, the correct option is (A).

Let P be any point on the circle with center O 

\({OP}=\) radius

Take a line L through \({P}\) and \({Q}\) as shown if \({L}\) is perpendicular to \(OP\).

\({OQ} \perp {OP}\) because the perpendicular distance is shortest. 

Every point except P lies outside the circle and line 1 must be a tangent. At any given point one and only one tangent can be drawn.

Thus, there is only one tangent at a point of the circle. 

Hence, the correct option is (A).