# Let the lengths of intercepts on

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 $\mathrm{x}+2 \mathrm{y}=0$, is euqal to :

1. (1) $\sqrt{11}$

2. (2) $\sqrt{7}$

3. (3) $\sqrt{6}$

4. (4) $\sqrt{10}$

Correct Option: , 3

Solution:

$x^{2}+y^{2}+a x+2 a y+c=0$

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

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

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

$\Rightarrow a^{2}-c=5$

(1)\& (2)

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

$\therefore \quad c=-1$

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

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

$\mathrm{m}_{\text {tangent }}=2$

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