Let the circles $$C_1 : |z| = r$$ and $$C_2 : |z - 3 - 4i| = 5$$, $$z \in \mathbb{C}$$, be such that $$C_2$$ lies within $$C_1$$. If $$z_1$$ moves on $$C_1$$, $$z_2$$ moves on $$C_2$$ and $$\min|z_1 - z_2| = 2$$, then $$\max|z_1 - z_2|$$ is equal to :

The two circles are
  • $$C_1 : |z| = r$$   (centre $$O = 0$$, radius $$r$$)
  • $$C_2 : |z - (3+4i)| = 5$$   (centre $$A = 3+4i$$, radius $$5$$)

Distance between the centres is
$$OA = |3 + 4i| = \sqrt{3^{2}+4^{2}} = 5 \; .$$ Hence $$d = 5$$.

Because $$C_2$$ lies completely inside $$C_1$$,
$$r \ge d + 5 = 10 \; . \quad -(1)$$

For two concentric or non-concentric circles (one inside the other) with radii $$r_1$$ (outer), $$r_2$$ (inner) and centre distance $$d$$, the shortest distance between points on the two circles is
$$\min |z_1 - z_2| = r_1 - (d + r_2). \quad -(2)$$

Here $$r_1 = r$$, $$r_2 = 5$$, $$d = 5$$ and the given minimum distance equals $$2$$:
$$r - (5 + 5) = 2 \; \Longrightarrow \; r - 10 = 2 \; \Longrightarrow \; r = 12. \quad -(3)$$

The greatest distance between points on the two circles is obtained when one point is on the outer circle farthest from the inner centre and the other point is on the inner circle farthest from the outer centre along the same line. The formula is
$$\max |z_1 - z_2| = r_1 + (d + r_2). \quad -(4)$$

Substituting $$r_1 = 12$$, $$d = 5$$, $$r_2 = 5$$ into $$(4)$$:
$$\max |z_1 - z_2| = 12 + (5 + 5) = 12 + 10 = 22.$$(5)

Therefore, the required maximum distance is $$22$$.

Option C which is: $$22$$