Question

For all complex numbers $z _{1}, z _{2}$ satisfying $\left| z _{1}\right|=12$ and $\left| z _{2}-(3+4 i )\right|=5$, the minimum value of $\left|z_{1}-z_{2}\right|$ is

• A. 4
• B. 3
• C. 1
• D. 2

Answer 1

Answer: (D)

2

Answer 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 ie, the centre of $C_{1}$. Now, $C_{1} C_{2}=\sqrt{3^{2}+4^{2}}=5$ and $r _{1}- r _{2}=125=7$ $\therefore C _{1} C _{2}< r _{1}- r _{2}$ Hence, circle $C _{2}$ lies inside the circle $C _{1}$. From figure the, minimum distance between them is $AB = C _{1} BC _{1} A = r _{1}-2 r _{2}$ $=12-10=2$