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

Answer 1

Answer: (C)

$x^2+y^2+ax+2ay+c=0$
$2\sqrt{g^2-c}=2\sqrt{\frac{a^2}{4}-c}=2\sqrt{2}$
$\Rightarrow \frac{a^2}{4}-c=2\dots$(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)
$(1)\&(2)$
$\frac{3a^2}{4}=3\Rightarrow a=-2(a<0)$
$\therefore c=-1$
Circle $\Rightarrow x^2+y^2-2x-4y-1=0$
$\Rightarrow (x-1{)}^2+(y-2{)}^2=6$
Given $x+2y=0\Rightarrow m=-\frac{1}{2}$
$m_{\text{tangent }}=2$
Equation of tangent
$\Rightarrow (y-2)=2(x-1)\pm \sqrt{6}\sqrt{1+4}$
$\Rightarrow 2x-y\pm \sqrt{30}=0$
Perpendicular distance from $(0,0)=\left|\frac{\pm \sqrt{30}}{\sqrt{4+1}}\right|=\sqrt{6}$