A point P lies on the circle x^{2}+y^{2}=169. | vedclass

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(B) The given equation of the circle is $x^{2}+y^{2}=169$. Its center is $O=(0, 0)$ and radius is $r=13$. Points $Q=(5, 12)$ and $R=(-12, 5)$ lie on t

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$\frac{\pi}{4}$

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(B) The given equation of the circle is $x^{2}+y^{2}=169$. Its center is $O=(0, 0)$ and radius is $r=13$. Points $Q=(5, 12)$ and $R=(-12, 5)$ lie on the circle since $5^{2}+12^{2}=169$ and $(-12)^{2}+5^{2}=169$. The slope of $OQ$ is $m_{1} = \frac{12-0}{5-0} = \frac{12}{5}$. The slope of $OR$ is $m_{2} = \frac{5-0}{-12-0} = -\frac{5}{12}$. Since $m_{1} \cdot m_{2} = \left(\frac{12}{5}\right) \cdot \left(-\frac{5}{12}\right) = -1$, the lines $OQ$ and $OR$ are perpendicular. Therefore, the central angle $\angle ROQ = \frac{\pi}{2}$. By the circle theorem, the angle subtended by a chord at the circumference is half the angle subtended by it at the center. Thus, $\angle QPR = \frac{1}{2} \angle ROQ = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$.

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

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