Question

If the line $ax+by=0$ touches the circle $x^2+y^2+2x+4y=0$ and is a normal to the circle $x^2+y^2-4x+2y-3=0,$ then value of $(a,b)$ is/are

• A. $(1,2)$
• B. $(1,-2)$
• C. $(-1,2)$
• D. $(-1,-2)$

Answer 1

Answer: (A), (D)

$(-1,-2)$

As the line $ax+by=0$ touches the circle $x^2+y^2+2x+4y=0,$ so the distance of the centre $(-1,-2)$ from the line is equal to radius
$\left|\frac{-a-2b}{\sqrt{a^2+b^2}}\right|=\sqrt{(-1{)}^2+(-2{)}^2-0}$
Squaring both the sides, we get
$\Rightarrow (a+2b{)}^2=5(a^2+b^2)$
$\Rightarrow 4a^2-4ab+b^2=0$
$\Rightarrow (2a-b{)}^2=0$
$\Rightarrow b=2a$

Now, $ax+by=0$ is a normal to the circle $x^2+y^2-4x+2y-3=0$, so it should pass through the centre $(2,-1)$
$\Rightarrow 2a-b=0$
$\Rightarrow b=2a$
Only, $(a,b)=(1,2)$ and $(-1,-2)$ are satisfying the condition.