Question

The tangents are drawn at the extremities of a diameter $AB$ of a circle with center $P$. If a tangent to the circle at the point $C$ intersects the other two tangents at $Q$ and $R$, then the measure of the $\angle QPR$ is:

• A. ${45}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. ${180}^o$

Answer 1

The correct option is C ${90}^o$

Construction:- Join PC, PQ and PR

we know that tangent is perpendicular to the radius through the point of contact

so, $PM\perp MQ,PC\perp QR$ and $PN\perp NR$

Now, In $\Delta PMQ$ and $\Delta PCQ$

$\angle PMQ=\angle PCQ$ each ${90}^0$

$PQ=PQ$ common

$PM=PC$ radii of the same circle

so, $\Delta PMQ\cong \Delta PCQ$ (by R.H.S)

$\Rightarrow \angle PQC=\frac{1}{2}\angle MQR$

$\Rightarrow \angle MQR=2\angle PQC-\left(1\right)$

Similarly, $\Delta PCR$ and $\Delta PNR$

$PR=PR$ common

$\angle PCR=\angle PNR$ each ${90}^0$

$PC=PN$ radii of the same circle

so

$\Delta PCR\cong \Delta PNR$

$\Rightarrow \angle PRC=\angle PRN$ by R.HS

$\Rightarrow \angle PRC=\angle PRN$

$\Rightarrow \angle PRC\frac{1}{2}\angle NRC-\left(2\right)$

NOW,

$\angle MQR+\angle NRQ={180}^0$ sum of exterior angle

$\Rightarrow 2\angle PQC+2\angle PRC={180}^0$ from equation 1 & 2

$\Rightarrow \angle PQC+\angle PRC={90}^0$

$\Rightarrow \angle PQR+\angle PRQ={90}^0$

In $\Delta PQR$

$\Rightarrow \angle PQR+\angle QPR+\angle QRP={180}^0$ by an angle sum property

$\Rightarrow \angle QPR+{90}^0={180}^0$

$\Rightarrow \angle QPR={90}^0$