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

Answer: (B)

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

A line $y=mx+c$ is tangent to a circle with radius ${}^{\prime }{a}^{\prime }$ if

$\Rightarrow c=\pm a\sqrt{m^2+1}$

Here, $y=\frac{-ax}{b}+\frac{2}{b}$

$\Rightarrow \frac{2}{b}=\pm 1\sqrt{\frac{a^2}{b^2}+1}$

$\Rightarrow \frac{2}{b}=\pm \frac{\sqrt{a^2+b^2}}{b}$

Squaring

$\Rightarrow a^2+b^2=4\to \left(1\right)$

If $ax+by=2$ is normal to $x^2+y^2-4x-4y=0$, it must pass through centre of the circle i.e. $(2,2)$

$\Rightarrow 2a+2b=2$

$\Rightarrow a+b=1\to \left(2\right)$

Putting $b=1-a$ in $\left(1\right)$

$\Rightarrow a^2+(1-a{)}^2=4$

$\Rightarrow a^2+1+a^2-2a=4$

$\Rightarrow 2a^2-2a=3$

$\Rightarrow 2a^2-2a-3=0$

$\Rightarrow a=\frac{2\pm \sqrt{4+24}}{4}$

$=\frac{2\pm \sqrt{28}}{4}$

$=\frac{2\pm 2\sqrt{7}}{4}$

$=\frac{1\pm \sqrt{7}}{2}$

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