Question

In triangle $PQR$ is inscribed in the circle ${{x}^{2}}+{{y}^{2}}=25$. If $Q$ and $R$ have coordinates $(3,4)$ and $(-4,3)$ respectively, then $\angle QPR$ is equal to
A. $\dfrac{\pi }{2}$
B. $\dfrac{\pi }{3}$
C. $\dfrac{\pi }{4}$
D. $\dfrac{\pi }{6}$

Answer 1

Hint: In this question, we are to find the angle at a vertex in the given triangle. To find this the slope with two points formula is used.

Formula used: The slope of a line that is passing through \[({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}})\] is
$m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
The product of the slopes of two lines if they are perpendicular is
${{m}_{1}}\times {{m}_{2}}=-1$

Complete step by step solution: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}=25$
This is in the form of ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ whose center lies at $O(0,0)$origin.
Then, the center of the given circle is $O(0,0)$
It is given that, $Q$ and $R$ have coordinates $(3,4)$ and $(-4,3)$ respectively.
Then, the slope of $\overleftrightarrow{OQ}$ is
$\begin{align}
  & {{m}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\
 & \text{ }=\dfrac{4-0}{3-0} \\
 & \text{ }=\dfrac{4}{3} \\
\end{align}$
And the slope of $\overleftrightarrow{OR}$ is
$\begin{align}
  & {{m}_{2}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\
 & \text{ }=\dfrac{3-0}{-4-0} \\
 & \text{ }=\dfrac{-3}{4} \\
\end{align}$
Then the product of these two slopes results as
$\begin{align}
  & {{m}_{1}}\times {{m}_{2}}=\dfrac{4}{3}\times \dfrac{-3}{4} \\
 & \text{ }=-1 \\
\end{align}$
Thus, these two lines are perpendicular. So, $\angle QOR=\dfrac{\pi }{2}$
Then,
 $\begin{align}
  & \angle QPR=\dfrac{1}{2}\times \angle QOR \\
 & \text{ }=\dfrac{1}{2}\times \dfrac{\pi }{2} \\
 & \text{ }=\dfrac{\pi }{4} \\
\end{align}$

Thus, Option (C) is correct.

Note: Here we need to calculate the slope. Thus, we can able to calculate the required angle. Since the product of two lines is $-1$, those lines are perpendicular to each other. So, we can write the angle between them is $\dfrac{\pi }{2}$.