Question

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

Answer 1

Answer: (C)

$\frac{\pi }{4}$

The given equation of circle is $x^2+y^2=25$

$\therefore$ Centre is $(0,0)$ and $r=5$

Now,

$OR=\sqrt{(-4-3{)}^2+(3-4{)}^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}$ units

$OQ=OR=5$ units

$\cos \angle QOR=\frac{O{Q}^2+O{R}^2-Q{R}^2}{2OQ\times OR}$

$\Rightarrow \cos \angle QOR=\frac{25+25-50}{2\times 5\times 5}=0$

$\Rightarrow \angle QOR={90}^o$

We know the angle subtended by an arc at the centre is double the angle subtended by same arc at any point on the circle.

Now, $\angle QOR=2\angle QPR$

$\Rightarrow {90}^o=2\angle QPR$

$\Rightarrow \angle QPR={45}^o=\frac{\pi }{4}$.