JEE maths#21

$|z_{1}|^{2} \le 169$ describes a disk $D_{1}$ in the complex plane centered at the origin with radius $13$ .

$|z_{2} + 3 - 4i|^{2} \le 25$ describes a disk $D_{2}$ centered at $(-3, 4)$ with radius $5$ , the circumference of which passes through the origin.

Now $D_{2}$ lies entirely within $D_{1}$ , so since $D_{1}$ is centered at the origin and the circumference of $D_{2}$ intersects the origin, the maximum distance between a point on $D_{1}$ and a point on $D_{2}$ will be the sum of the radius of $D_{1}$ and the diameter of $D_{2}$ , i.e., $13 + 2 \times 5 = \boxed{23}$ .

Comment: The most distant points involved are $\left(\dfrac{39}{5}, -\dfrac{52}{5}\right)$ on $D_{1}$ and $(-6, 8)$ on $D_{2}$ .

As $|z_1|$ is $+ve$ , hence we can rewrite the equation one as

$|z_1| \leq 13$ , Similarly $|z_2+3-4i| \leq 5$

Now $max(z_1)=13$ .

Again $|z_2+3-4i| \leq 5$

Or $|z_2|- 5 \leq 5$ or $|z_2| \leq 10.$

Now Max of $|z_2|$ is $10$ .

But we need $|z_1 - z_2| = |z_1 + (-z_2)| \leq |z_1|-|z_2|$

And Hence Maximum of |z 1 - z 2| = $13- (-10) = 23$