Question

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

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (B)

Given equation of circle is
$x^2+y^2=169$
Its centre $=(0,0)$ and radius $=13$

Now, slope of $OR=\frac{-5}{12}=m_1(\because$ slope $=\frac{y_2-y}{x_2-x})$
and slope of $OQ=\frac{12}{5}=m_2$

$\because m_1\cdot m_2=-1$
$\Rightarrow \angle ROQ=\frac{\pi }{2}$
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.
$\therefore \angle QPR=\frac{1}{2}\cdot \angle ROQ=\frac{1}{2}\times \frac{\pi }{2}=\frac{\pi }{4}$