Question

Let the lengths of intercepts on $x$-axis and $y$-axis made by the circle $x^2+y^2+ax+2ay+c=0,$ $(a<0)$ be $2\sqrt{2}$ and $2\sqrt{5},$ respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line $x+2y=0,$ is equal to :

• A. $\sqrt{10}$
• B. $\sqrt{6}$
• C. $\sqrt{11}$
• D. $\sqrt{7}$

Answer 1

The correct option is B $\sqrt{6}$

$2\sqrt{\frac{a^2}{4}-c}=2\sqrt{2}$
$\Rightarrow \sqrt{a^2-4c}=2\sqrt{2}$
$\Rightarrow a^2-4c=8...\left(1\right)$

$2\sqrt{a^2-c}=2\sqrt{5}$
$\Rightarrow a^2-c=5...\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right),$
$3c=-3\Rightarrow c=-1$
$a^2=4\Rightarrow a=-2$
So, the circle is $x^2+y^2-2x-4y-1=0$

Equation of tangent : $2x-y+\lambda =0$
$\therefore$ Perpendicular distance from centre to tangent $=$ radius
$\left|\frac{2-2+\lambda }{\sqrt{5}}\right|=\sqrt{6}$
$\Rightarrow \lambda =\pm \sqrt{30}$
$\therefore$ Tangents are $2x-y\pm \sqrt{30}=0$
Distance from origin $=\frac{\sqrt{30}}{\sqrt{5}}=\sqrt{6}$