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Let the lengths of intercepts on $x$ -axis and $y$ -axis made by the circle $x^{2}+y^{2}+a x+2 a y+c=0$ $(a < 0)$ 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 $x +2 y =0,$ is euqal to :
$\sqrt{11}$
$\sqrt{7}$
$\sqrt{6}$
$\sqrt{10}$
Solution
$x ^{2}+ y ^{2}+ ax +2 ay + c =0$
$2 \sqrt{ g ^{2}- c }=2 \sqrt{\frac{ a ^{2}}{4}- c }=2 \sqrt{2}$
$\Rightarrow \quad \frac{ a ^{2}}{4}- c =2…….(1)$
$2 \sqrt{ f ^{2}- c }=2 \sqrt{ a ^{2}- c }=2 \sqrt{5}$
$\Rightarrow a^{2}-c=5……(2)$
$(1)$ and $(2)$
$\frac{3 a ^{2}}{4}=3 \Rightarrow a =-2 \quad( a < 0)$
$\therefore \quad c=-1$
Circle $\Rightarrow x^{2}+y^{2}-2 x-4 y-1=0$
$\Rightarrow(x-1)^{2}+(y-2)^{2}=6$
Given $x+2 y=0 \Rightarrow m=-\frac{1}{2}$
$m _{\text {tangent }}=2$
Equation of tangent
$\Rightarrow(y-2)=2(x-1) \pm \sqrt{6} \sqrt{1+4}$
$\Rightarrow 2 x-y \pm \sqrt{30}=0$
Perpendicular distance from $(0,0)=\left|\frac{\pm \sqrt{30}}{\sqrt{4+1}}\right|=\sqrt{6}$
