Question

The angle between tangents drawn from the point $(-1,3)$ to the circle $x^2+y^2=5$ is

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

Answer 1

Answer: (C)

$\frac{\pi }{2}$

Given:Circle:$x^2+y^2={\left(\sqrt{5}\right)}^2$ where radius of the circle $r=\sqrt{5}$ units

A tangent with slope $m$ is $y=mx\pm r\sqrt{1+m^2}=mx\pm \sqrt{5}\sqrt{1+m^2}$

It passes through $(-1,3)$

$\therefore 3=-m\pm \sqrt{5}\sqrt{1+m^2}$

$\Rightarrow m+3=\pm \sqrt{5}\sqrt{1+m^2}$

Squaring both sides,we get

$\Rightarrow {(m+3)}^2=5(1+m^2)$

$\Rightarrow m^2+6m+9-5-5m^2=0$

$\Rightarrow -4m^2+6m-4=0$

$\Rightarrow -2(2m^2-3m-2)=0$

$\Rightarrow 2m^2-3m-2=0$

$\Rightarrow 2m^2-4m+m-2=0$

$\Rightarrow 2m(m-2)+(m-2)=0$

$\Rightarrow (m-2)(2m+1)=0$

$\Rightarrow m_1=2,m_2=-\frac{1}{2}$

If $m_1$ and $m_2$ are the roots, then $m_1{m}_2=-1$

$\Rightarrow 2\times \frac{-1}{2}=-1$

$\therefore$ the tangents are perpendicular.

Hence angle made between the tangents is $\frac{\pi }{2}$