The triangle $P Q R$ is inscribed in the circle $x^2+y^2=25$. If $\mathrm{Q}=(3,4)$ and $\mathrm{R}=(-4,3)$…
The triangle $P Q R$ is inscribed in the circle $x^2+y^2=25$. If $\mathrm{Q}=(3,4)$ and $\mathrm{R}=(-4,3)$ then $\angle \mathrm{QPR}=$
$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$
Solution
The given equation of circle is $x^2+y^2=25$
$\therefore$ Centre is $(0,0)$ and $r=5$
Now, QR
$\begin{aligned}
& =\sqrt{(-4-3)^2+(3-4)^2}=5 \sqrt{2} \\
& O Q=O R=5
\end{aligned}$
$\begin{aligned}
& \cos (\angle \mathrm{QOR})=\frac{\mathrm{OQ}^2+\mathrm{OR}^2-\mathrm{QR}^2}{2 \mathrm{OQ} \times \mathrm{OR}}=\frac{25+25-50}{2 \times 5 \times 5}=0 \\
& \Rightarrow \angle \mathrm{QOR}=\frac{\pi}{2}
\end{aligned}$ We know the angle subtended by an arc at the centre is double the angle subtended by same arc at any point on the circle.
Now, $\angle \mathrm{QOR}=2 \angle \mathrm{QPR}$
$\Rightarrow \frac{\pi}{2}=2 \angle \mathrm{QRP} \Rightarrow \angle \mathrm{QPR}=\frac{\pi}{4}$
