Question

If the line $ax+by=2$ is a normal to the circle $x^2+y^2+4x+4y=0$ and a tangent to the circle $x^2+y^2=1$, then

• A. $a=\frac{1}{2},b=\frac{1}{2}$
• B. $a=\frac{-1+\sqrt{7}}{2},b=\frac{-1-\sqrt{7}}{2}$
• C. $a=\frac{1}{4},b=\frac{3}{4}$
• D. $a=1,b=\sqrt{3}$

Answer 1

The correct option is B $a=\frac{-1+\sqrt{7}}{2},b=\frac{-1-\sqrt{7}}{2}$

$S_1:x^2+y^2+4x+4y=0$

$S_2:x^2+y^2=1$

The normal of a circle $S_1$ passes through its centre.

So, $ax+by=2$ passes through the centre $(-2,-2)$
$\Rightarrow -2a-2b=2$
$\Rightarrow a+b=-1$

Distance from origin to the tangent of the circle $S_2$ is equal to radius.

So, the distance from $(0,0)$ to the $ax+by=2$ is radius i.e $1$.
$\Rightarrow \frac{2}{\sqrt{a^2+b^2}}=1$
$\Rightarrow a^2+b^2=4$
$\Rightarrow a^2+(-1-a{)}^2=4$
$\Rightarrow 2a^2+2a-3=0$
$\Rightarrow a=\frac{-1+\sqrt{7}}{2},\frac{-1-\sqrt{7}}{2}$

For $a=\frac{-1+\sqrt{7}}{2},b=\frac{-1-\sqrt{7}}{2}$

For $a=\frac{-1-\sqrt{7}}{2},b\frac{-1+\sqrt{7}}{2}$

Hence, option B.