Mathematics - Circle Question with Solution | TestHub

$x^2+y^2+ax+2ay+c=0$

$2\sqrt{g^2-c}=2\sqrt{\frac{a^2}{4}-c}=2\sqrt{2}$

$\Rightarrow \frac{a^2}{4}-c=2 ...\left(1\right)$

$2\sqrt{f^2-c}=2\sqrt{a^2-c}=2\sqrt{5}$

$\Rightarrow a^2-c=5 ...\left(2\right)$

$\left(1\right) \& \left(2\right)$

$\frac{3a^2}{4}=3\Rightarrow a=-2 \left(a<0\right)$

$\therefore c=-1$

Circle $\Rightarrow x^2+y^2-2x-4y-1=0$

$\Rightarrow {\left(x-1\right)}^2+{\left(y-2\right)}^2=6$

Given $x+2y=0\Rightarrow m=-\frac{1}{2}$

$m_{\mathrm{tangent}}=2$

Equation of tangent $\Rightarrow \left(y-2\right)=2\left(x-1\right)\pm \sqrt{6}\sqrt{1+4}$

$\Rightarrow 2x-y\pm \sqrt{30}=0$

Perpendicular distance from $\left(0,0\right)=\left|\frac{\pm \sqrt{30}}{\sqrt{4+1}}\right|=\sqrt{6}$