Question

The equation of a circle of radius $2$ touching the circles $x^2+y^2-4\left|x\right|=0$ is

• A. $x^2+y^2+2\sqrt{3}y+2=0$
• B. $x^2+y^2+4\sqrt{3}y+8=0$
• C. $x^2+y^2-4\sqrt{3}y+8=0$
• D. None of these

Answer 1

Answer: (B), (C)

The correct options are
B $x^2+y^2+4\sqrt{3}y+8=0$
C $x^2+y^2-4\sqrt{3}y+8=0$

The given circle are

$x^2+y^2-4x=0,x>0$

i.e., ${(x-2)}^2+y^2=2^2,x>0.$

and, $x^2+y^2+4x=0,x<0$

i.e., ${(x+2)}^2+y^2=2^2,x<0.$

So, the centres of the given circles are $(-2,0)$ and $(2,0)$ and both having same radius $2$ units.

Clearly, from the figure, the centers of the circles forms an equilateral triangle of side $4$ units.

Height of this equilateral triangle $=\frac{\sqrt{3}}{2}\times 4=2\sqrt{3}=\sqrt{12}$

So, the centres of the required circle are at $(0,\sqrt{12})$ and $(0,-\sqrt{12}).$

$\therefore$ Equations of the required circle are

${(x-0)}^2+{(y\mp \sqrt{12})}^2=2^2$

i.e., $x^2+y^2+4\sqrt{3}y+8=0.$

and $x^2+y^2-4\sqrt{3}y+8=0.$