Question

In the given figure, R is an external point and RP and RQ are tangents drawn to the circle from R. $\angle POQ$ is ${90}^{\circ }$. The quadrilateral OPRQ is a ____ and the $\angle POR$ is ____

• A. $\text{Kite},{60}^{\circ }$
• B. $\text{Square},{60}^{\circ }$
• C. $\text{Rhombus},{45}^{\circ }$
• D. $\text{Square},{45}^{\circ }$

Answer 1

The correct option is D $\text{Square},{45}^{\circ }$


Given that $\angle POQ$ = ${90}^{\circ }$
A tangent to the circle is perpendicular to the radius at the point of contact
Hence, $\angle OPR={90}^{\circ }and\angle OQR={90}^{\circ }$
Now, in the quadrilateral OPRQ,
$\angle POQ+\angle OQR+\angle QRP+\angle RPO={360}^{\circ }.$
${90}^{\circ }+{90}^{\circ }+\angle QRP+{90}^{\circ }={360}^{\circ }$
Hence, $\angle QRP={90}^{\circ }.$
Since the angles are ${90}^{\circ }$, OP is parallel to RQ and OQ is parallel to RP.
Hence, OPQR is a parallelogram, and OQ = RP and OP = RQ;
Now, PR = RQ, since the lengths of tangents drawn from an external point to a circle are equal in length. And OP = OQ as they are radii of the same circle
Hence, OP = PR = RQ = QO. And all angles are ${90}^{\circ }$.
Hence, OPRQ is a square
Now, OR bisects $\angle$ PRQ since the line drawn from the center of the circle to an external point bisects the angle between the tangents drawn from that point.
Hence, $\angle ORP=\angle ORQ=\frac{{90}^{\circ }}{2}={45}^{\circ }.$
In the triangle OPR,
$\angle ORP+\angle POR+\angle OPR={180}^{\circ }$
${45}^{\circ }+{90}^{\circ }+\angle POR={180}^{\circ }$
Hence, $\angle POR={45}^{\circ }$