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A point $P$ lies on the circle $x^{2}+y^{2}=169$. If $Q=(5,12)$ and $R=(-12,5),$ then the $\angle Q P R$ is
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The correct answer is:
$\frac{\pi}{4}$
Given equation of circle is
$$
x^{2}+y^{2}=169
$$
Its centre $=(0,0)$ and radius $=13$
Now, slope of $O R=\frac{-5}{12}=m_{1}\left(\because\right.$ slope $\left.=\frac{y_{2}-y}{x_{2}-x}\right)$
and slope of $O Q=\frac{12}{5}=m_{2}$
$\because m_{1} \cdot m_{2}=-1 \Rightarrow \angle R O Q=\frac{\pi}{2}$
We know that, angle made by the chord of circle at circumference is equal to the half of the angle made by the same chord at the centre of circle.
$\therefore \quad \angle Q P R=\frac{1}{2} \cdot \angle R O Q=\frac{1}{2} \times \frac{\pi}{2}=\frac{\pi}{4}$
$$
x^{2}+y^{2}=169
$$
Its centre $=(0,0)$ and radius $=13$
Now, slope of $O R=\frac{-5}{12}=m_{1}\left(\because\right.$ slope $\left.=\frac{y_{2}-y}{x_{2}-x}\right)$
and slope of $O Q=\frac{12}{5}=m_{2}$
$\because m_{1} \cdot m_{2}=-1 \Rightarrow \angle R O Q=\frac{\pi}{2}$
We know that, angle made by the chord of circle at circumference is equal to the half of the angle made by the same chord at the centre of circle.
$\therefore \quad \angle Q P R=\frac{1}{2} \cdot \angle R O Q=\frac{1}{2} \times \frac{\pi}{2}=\frac{\pi}{4}$
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