Question

The equation(s) of circle(s) of radius $1$ touching the circle $x^2+y^2-2\left|x\right|=0$ is/are

• A. $x^2+y^2+2\sqrt{3}x-2=0$
• B. $x^2+y^2-2\sqrt{3}y+2=0$
• C. $x^2+y^2+2\sqrt{3}y+2=0$
• D. $x^2+y^2+2\sqrt{3}x+2=0$

Answer 1

Answer: (B), (C)

$x^2+y^2+2\sqrt{3}y+2=0$
$x^2+y^2-2\sqrt{3}y+2=0$

Given equation of circle is
$x^2+y^2-2|x|=0$
This equation can be simplified as :
$x^2+y^2-2x=0$ , ($x>0$) and
$x^2+y^2+2x=0$, ($x<0$)
$\Rightarrow (x-1{)}^2+(y-0{)}^2-1=0$ and $(x+1{)}^2+(y-0{)}^2-1=0$
So, we have circles having centers at $(1,0)$ and $(-1,0)$ with radius $1$
Now, the required circle touches these two circles.
By symmetry,the centre of the required circle will lie on the y-axis. Hence, the coordinates of the centre will be $(0,b)$
So, distance between the centres $(0,b)$ and $(1,0)$ $=r_1+r_2$
$\Rightarrow b^2+1=4$
$\Rightarrow b=\pm \sqrt{3}$
Hence there would be two circles which touch the given set of circles externally.
$x^2+(y\pm \sqrt{3}{)}^2=1$
This can be simplified as -
$x^2+y^2-2\sqrt{3}y+2=0$
$x^2+y^2+2\sqrt{3}y+2=0$
Hence, options 'B' and 'C' are correct.