Question

For all complex numbers $z_1$ and $z_2$ satisfying $\left|z_1\right|=12$ and $|z_2-3-4i|=5$, the minimum value of $|z_1-z_2|$ is

• A. $0$
• B. $2$
• C. $7$
• D. $17$

Answer 1

Answer: (B)

$2$

The two circles whose centre and radius are $C_1(0,0),r_1=12,C_2(3,4),r_2=5$ and it passes through origin

Then we have
$C_1{C}_2=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{3^2+4^2}=5$

and $r_1-r_2=12-5=7$

$\therefore C_1{C}_2<r_1-r_2$

Hence, circle $C_2$ lies inside circle $C_1$

Now, the minimum distance between them is

$AB=C_{1}B-C_{1}A=r_1-2r_2=12-10=2$

$\therefore |z_1-z_2|=2$