The $\triangle P Q R$ is inscribed in the circle $x^2+y^2=25$. If $Q$ and $R$ have coordinates $(3,4)$ and $(-4,3)$, respectively. Then, $\angle Q P R$ is equal to

• A

  $\frac{\pi}{2}$
• B

  $\frac{\pi}{3}$
• Correct answer
  C

  $\frac{\pi}{4}$
• D

  $\frac{\pi}{6}$

Solution

Let $O$ is the point on centre and $P$ is the point on circumference. Therefore, angle QOR is double the angle QPR. So, it is sufficient to find the angle $Q O R$.
Now, slope of $O Q, m_1=\frac{4-0}{3-0}=\frac{4}{3}$ Slope of $O R, m_2=\frac{3-0}{-4-0}=-\frac{3}{4}$ Again, $\quad m_1 m_2=-1$ Therefore, $\quad \angle Q O R=90^{\circ}$ Which implies that $\angle Q P R=\frac{90^{\circ}}{2}=45^{\circ}$.

Hence, the correct answer is option: 'C'.