Question

If the angle between the tangents drawn to $x^2+y^2+4x+8y+c=0$ from $(0,0)$ is $\frac{\pi }{2}$ ,then find the value of ${}^{\prime }{c}^{\prime }$

Answer 1

From the above equation $2gx=4x\Rightarrow g=2$ and $2fy=8y\Rightarrow f=4$

Let $y=mx$ be the tangent perpendicular from centre $(-2,-4)$

on the tangent$=$ radius of the circle.

So,$\frac{-2m+4}{\sqrt{1+m^2}}=\sqrt{4+16-c}$

${\left(\frac{-2m+4}{\sqrt{1+m^2}}\right)}^2=20-c$

$\Rightarrow 4m^2+16-16m=(20-c)(1+m^2)$

$\Rightarrow 4m^2+16-16m=20+20m^2-c-c{m}^2$

$\Rightarrow -16m^2+c{m}^2-16m+(c-4)=0$

$\Rightarrow (c-16)m^2-16m+(c-4)=0$ is quadratic in $m$

$\therefore m=\frac{16\pm \sqrt{256-4(c-16)(c-4)}}{2(c-16)}$

$m=\frac{16\pm 2\sqrt{64-c^2+24c-64}}{2(c-16)}$

$=\frac{8\pm \sqrt{24c-c^2}}{(c-16)}$

$\therefore m_1=\frac{8+\sqrt{24c-c^2}}{(c-16)}$ and $m_2=\frac{8-\sqrt{24c-c^2}}{(c-16)}$

Given $m_1{m}_2=-1$

$\Rightarrow \frac{8+\sqrt{24c-c^2}}{(c-16)}\times \frac{8-\sqrt{24c-c^2}}{(c-16)}=-1$

$\Rightarrow \frac{64-{\left(\sqrt{24c-c^2}\right)}^2}{{(c-16)}^2}=-1$

$\Rightarrow 64-{\left(\sqrt{24c-c^2}\right)}^2=-{(c-16)}^2$

$\Rightarrow c^2-24c+64=-c^2-256+32c$

$\Rightarrow 2c^2-56c+320=0$ on simplification

$\Rightarrow c^2-28c+160=0$ by dividing by $2$

$\Rightarrow (c-20)(c-8)=0$ on factorising

$\Rightarrow c=20,8$