Question

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

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

Answer 1

Answer: (C)

$(1,2)$

Case I:-

Given that the line $ax+by=0$ touches the circle $x^2+y^2+2x+4y=0$, i.e., the line is tangent to the circle.

Equation of circle can be written as-

${(x+1)}^2+{(y+2)}^2-1-4=0$

$\Rightarrow {(x+1)}^2+{(y+2)}^2={\left(\sqrt{5}\right)}^2$

Therefore,

Centre of circle $=(-1,-2)$

Radius of circle $=\sqrt{5}$

As we know that perpendicular distance from a point $(x_1,y_1)$ to the line $ax+by+c=0$ is given by-

$d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

$\therefore \perp$ distance from the centre of circle to the line $ax+by=0$-

$d=\frac{|-a-2b|}{\sqrt{a^2+b^2}}$

As we know that the perpendicular distance from the centre of circle to the line touching the circle is equal to the radius of circle.

$\therefore \frac{|-a-2b|}{\sqrt{a^2+b^2}}=\sqrt{5}$

Squaring both sides, we have

$\frac{a^2+4b^2+4ab}{a^2+b^2}=5$

$\Rightarrow a^2+4b^2+4ab=5a^2+5b^2$

$\Rightarrow 4a^2+b^2-4ab=0$

$\Rightarrow {(2a-b)}^2=0$

$\Rightarrow 2a-b=0.....\left(1\right)$

Case II:-

Given that the line $ax+by=0$ is normal to the circle $x^2+y^2-4x+2y-3=0$, i.e., the line passes through the centre of circle.

The equation of circle can also be written as-

${(x-2)}^2+{(y+1)}^2-4-1-3=0$

$\Rightarrow {(x-2)}^2+{(y+1)}^2={\left(2\sqrt{2}\right)}^2$

Centre of circle $=(2,-1)$

As the line passes through its centre, i.e., point $(2,-1)$ will satisfy the equation $ax+by=0$.

$\therefore 2a-b=0.....\left(2\right)$

From the given option, $\left(C\right)(1,2)$ clearly satisfies the equation $\left(1\right)\&\left(2\right)$.

Hence the required answer is $\left(C\right)(1,2)$.