Question

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 equal to :

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

Answer 1

Answer: (C)

$x^{2}+y^{2}+a x+2 a y+c=0$
$2 \sqrt{ g ^{2}- c }=2 \sqrt{\frac{ a ^{2}}{4}- c }=2 \sqrt{2}$
$\Rightarrow \frac{ a ^{2}}{4}- c =2 \dots$(i)
$2 \sqrt{ f ^{2}- c }=2 \sqrt{ a ^{2}- c }=2 $
$2 \sqrt{ f ^{2}- c }=2 \sqrt{ a ^{2}- c }=2 \sqrt{5}$
$\Rightarrow a ^{2}- c =5 \dots$(ii)
$(1) \& (2)$
$\frac{3 a^{2}}{4}=3 \Rightarrow a=-2 (a<0)$
$\therefore c=-1$
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 $(0,0)=\left|\frac{\pm \sqrt{30}} {\sqrt{4+1}}\right|=\sqrt{6}$