A point $P$ lies on the circle $x^{2} + y^{2} = 169$ . If $Q = (5, 12)$ and $R = (- 12, 5)$ , then the $∠ Q P R$ is

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is C $\frac{π}{4}$
Given equation of circle is
$x^{2} + y^{2} = 169$
Its centre $= (0, 0)$ and radius $= 13$
Now slope of $O R = \frac{- 5}{12} = m_{1}$ ( $∵$ Slope $= \frac{y_{2} - y}{x_{2} - x}$ )
and slope of $O Q = \frac{12}{5} = m_{2}$
$∵$ $m_{1} . m_{2} = - 1$ $\Rightarrow$ $∠ R O Q = \frac{π}{2}$
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.
$∴$ $∠ Q P R = \frac{1}{2} . ∠ R O Q = \frac{1}{2} \times \frac{π}{2} = \frac{π}{4}$

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