Let the lengths of intercepts on x-axis and
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 euqal to :

  1. $\sqrt{11}$

  2. $\sqrt{7}$

  3. $\sqrt{6}$

  4. $\sqrt{10}$


Correct Option: , 3

Solution:

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

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

$\Rightarrow \frac{\mathrm{a}^{2}}{4}-\mathrm{c}=2$……(1)

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

$\Rightarrow \quad a^{2}-c=5$ ………..(2)

(1) & (2)

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

$\therefore \quad c=-1$

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

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

Given $x+2 y=0 \Rightarrow m=-\frac{1}{2}$

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

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

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