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. $a=1,b=2$

Answer 1

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

$x^2+y^2-4x-4y=0$
Centre of the circle is $(2,2)$
Each normal of a circle passes through the centre of the circle.
So, $2a+2b=2$
$\Rightarrow a+b=1\cdots \left(1\right)$

Centre and radius of $x^2+y^2=1$ are $(0,0)$ and $1$ respectively.
$ax+by=2$ is a tangent to $x^2+y^2=1^2$
So, $\left|\frac{-2}{\sqrt{a^2+b^2}}\right|=1$
$\Rightarrow a^2+b^2=4\cdots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get
$a=\frac{1+\sqrt{7}}{2},b=\frac{1-\sqrt{7}}{2}$
or $a=\frac{1-\sqrt{7}}{2},b=\frac{1+\sqrt{7}}{2}$