Question

Each of the circles $|z-1-i|=1$ and $|z-1+i|=1$ where $z=x+iy,$ touches internally a circle of radius $2$ units. The equation of the circle touching all the three circles can be

• A. $3z\overline{z}+z+\overline{z}-1=0$
• B. $3z\overline{z}-7z-7\overline{z}+15=0$
• C. $z\overline{z}-z-\overline{z}-3=0$
• D. $3z\overline{z}+z+\overline{z}+1=0$

Answer 1

Answer: (A), (B)

The correct options are
A $3z\overline{z}+z+\overline{z}-1=0$
B $3z\overline{z}-7z-7\overline{z}+15=0$


$C_1:|z-1-i|=1$
Centre of $C_1$ is $(1,1)$ and radius is $1$
$C_2:|z-1+i|=1$
Centre of $C_2$ is $(1,-1)$ and radius is $1$
Circle $C$ with radius $2$
$C_1$ and $C_2$ with radius $1$ touch this circle $C$ internally.
Circle $C_3$ and $C_4$ touch all the three circles.

Using Pythagoras theorem,
$(1{)}^2+(2-r{)}^2=(1+r{)}^2$
$\Rightarrow 1+4-4r+r^2=1+2r+r^2$
$\Rightarrow r=\frac{2}{3}$

Centres of $C_3$ and $C_4$ are $(-1+\frac{2}{3},0)$ and $(3-\frac{2}{3},0)$ respectively.
i.e., Centres of $C_3$ and $C_4$ are $(\frac{-1}{3},0)$ and $(\frac{7}{3},0)$ respectively.

Required equation of circle is
$|z+\frac{1}{3}|=\frac{2}{3}$ or $|z-\frac{7}{3}|=\frac{2}{3}$