Question

From point $P(-1,-2)$ , $PQ$ and $PR$ are the tangents drawn to the circle $x^2+y^2-6x-8y=0$. Then angle subtended by $QR$ on the centre of circle is

• A. $\pi -2{\sin }^{-1}(\frac{5}{2\sqrt{13}})$
• B. $\pi -{\sin }^{-1}(\frac{5}{2\sqrt{13}})$
• C. ${\cos }^{-1}\left(\frac{5}{2\sqrt{13}}\right)$
• D. ${\cos }^{-1}\left(\frac{5}{\sqrt{13}}\right)$

Answer 1

Answer: (A)

$\pi -2{\sin }^{-1}(\frac{5}{2\sqrt{13}})$

$x^2+y^2-6x-8y=0$
Radius$=5$
$PQ=\sqrt{S_1}=\sqrt{27}=3\sqrt{3}$
$\tan \alpha =\frac{OQ}{PQ}=\frac{5}{3\sqrt{3}}$
$\therefore \angle QOR$ is the angle sub tended by QR on the centre of circle
$=180-2\alpha$
Now, $180-2\alpha =180-2{\tan }^{-1}\left(\frac{5}{3\sqrt{3}}\right)$
$={180}^{\circ }-2{\sin }^{-1}\left(\frac{5}{2\sqrt{13}}\right)$
$=\pi -2{\sin }^{-1}\left(\frac{5}{2\sqrt{13}}\right)$