Question

The locus of the centers of the circles which touch the two circle $x^2+y^2=a^2$ and $x^2+y^2=4ax$ externally is

• A. $12x^2-4y^2-24ax+9a^2=0$
• B. $12x^2+4y^2-24ax+9a^2=0$
• C. $12x^2-4y^2+24ax+9a^2=0$
• D. $12x^2+4y^2+24ax+9a^2=0$

Answer 1

The correct option is A $12x^2-4y^2-24ax+9a^2=0$

Let $x^2+y^2+2gx+2fy+c=0$ be the variable circle.

Since it touches the given circle externally

$\therefore \sqrt{{(-g-0)}^2+{(-f-0)}^2}=\sqrt{g^2+f^2-c}+a$ ...(1)

and, $\sqrt{{(-g-2a)}^2+{(-f-0)}^2}=\sqrt{g^2+f^2-c}+2a$ ...(2)

Subtracting (1) from (2), we get

$\sqrt{{(g+2a)}^2+f^2}=\sqrt{g^2+f^2}+a.$

Squaring both sides, we get

${(g+2a)}^2+f^2=a^2+g^2+f^2+2a\sqrt{g^2+f^2}$

$\Rightarrow 4ag+4a^2=a^2+2a\sqrt{g^2+f^2}$

$\Rightarrow {(4g+3a)}^2=4(g^2+f^2)$

or, ${(-4(-g)+3a)}^2=4[{\left(g\right)}^2+{\left(f\right)}^2].$

$\therefore$ Locus of centre $(-g,-f)$ is ${(-4x+3a)}^2=4(x^2+y^2)$

or, $12x^2-4y^2-24ax-9a^2=0.$