Question

Angle between tangent $x^2+y^2=20,$ drawn to circles from the point $(6,2)$ is

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

Answer 1

The correct option is A

$\frac{\pi }{2}$

Explanation for the correct option:

Step 1. The equation of the tangent to the circle $x^2+y^2=20$ is $y=mx\pm \sqrt{20}\left(1+m^2\right).$

This tangent passes through the point $\left(6,2\right)$

put $x=6$ and $y=2$

$\Rightarrow$ $2=6m\pm \sqrt{20}\left(1+m^2\right)$

$\Rightarrow$ $2-6m=\pm \sqrt{20}\left(1+m^2\right)$

Step 2. Take square on Both sides.

${\left(2-6m\right)}^2={\sqrt{20}}^2\left(1+m^2\right)$

$\Rightarrow$ ${\left(2-6m\right)}^2=20\left(1+m^2\right)$ $\left[\mathbf{\because }{\left(a-b\right)}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{a}\mathit{b}\right]$

$\Rightarrow$ $\left(4+36m^2-24m\right)=\left(20+20m^2\right)$

$\Rightarrow$$4+36m^2-24m-20-20m^2=0$

$\Rightarrow$ $-16+16m^2-24m=0$

$\Rightarrow$ $16m^2-24m-16=0$

Step 3. Divide above equation by $8.$

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

$\Rightarrow$ $2m^2-4m+1m-2=0$

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

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

$\Rightarrow$ $\left(2m+1\right)=0$

$\Rightarrow$ $2m=-1$

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

and $m-2=0$

$\Rightarrow$ $m=2$

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

Step 4. Required angle$={\tan }^{-1}\left(\frac{m_1+m_2}{1+m_1{m}_2}\right)$

$$\begin{gathered} ={\tan }^{-1}\left(\frac{2-{\displaystyle \frac{1}{2}}}{1+2{\displaystyle \frac{-1}{2}}}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{4-1}{2}}}{{\displaystyle \frac{2-4}{2}}}\right) \\ ={\tan }^{-1}\left(\frac{3\times 2}{-2\times 2}\right) \\ ={\tan }^{-1}\left(-\frac{3}{2}\right) \\ ={\tan }^{-1}\left(\infty \right) \\ \mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}} \end{gathered}$$

Hence, option(A) is correct