Question 1210364
Here's how to solve the problem:

Let $z_1$ and $z_2$ be two complex numbers.
We are given two conditions:
1.  $|z_1| = 5$
2.  $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 0$

From the second condition, we can multiply by $z_1 z_2$ (assuming $z_1 \neq 0$ and $z_2 \neq 0$, which must be true since $|z_1| = 5$ and if $z_2=0$ the expression is undefined):
$z_1^2 + z_2^2 = 0$
$z_1^2 = -z_2^2$

Now, let's take the modulus of both sides:
$|z_1^2| = |-z_2^2|$
$|z_1|^2 = |-1| |z_2|^2$
$|z_1|^2 = 1 \cdot |z_2|^2$
$|z_1|^2 = |z_2|^2$

Since $|z_1| = 5$, we have:
$5^2 = |z_2|^2$
$25 = |z_2|^2$
So, $|z_2| = 5$.

Now we need to find $|z_1 - z_2|^2$.
We know that for any complex number $z$, $|z|^2 = z \bar{z}$.
So, $|z_1 - z_2|^2 = (z_1 - z_2)(\overline{z_1 - z_2})$
$|z_1 - z_2|^2 = (z_1 - z_2)(\bar{z_1} - \bar{z_2})$
$|z_1 - z_2|^2 = z_1 \bar{z_1} - z_1 \bar{z_2} - z_2 \bar{z_1} + z_2 \bar{z_2}$
$|z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - (z_1 \bar{z_2} + z_2 \bar{z_1})$

We already know $|z_1|^2 = 25$ and $|z_2|^2 = 25$. So:
$|z_1 - z_2|^2 = 25 + 25 - (z_1 \bar{z_2} + z_2 \bar{z_1})$
$|z_1 - z_2|^2 = 50 - (z_1 \bar{z_2} + z_2 \bar{z_1})$

Let's go back to the relation $z_1^2 = -z_2^2$.
We can write $z_1 = i z_2$ or $z_1 = -i z_2$.
Case 1: $z_1 = i z_2$
Then $\bar{z_1} = -i \bar{z_2}$.
Substitute these into $z_1 \bar{z_2} + z_2 \bar{z_1}$:
$z_1 \bar{z_2} + z_2 \bar{z_1} = (i z_2) \bar{z_2} + z_2 (-i \bar{z_2})$
$= i |z_2|^2 - i |z_2|^2$
$= 0$

Case 2: $z_1 = -i z_2$
Then $\bar{z_1} = i \bar{z_2}$.
Substitute these into $z_1 \bar{z_2} + z_2 \bar{z_1}$:
$z_1 \bar{z_2} + z_2 \bar{z_1} = (-i z_2) \bar{z_2} + z_2 (i \bar{z_2})$
$= -i |z_2|^2 + i |z_2|^2$
$= 0$

In both cases, $z_1 \bar{z_2} + z_2 \bar{z_1} = 0$.

Therefore,
$|z_1 - z_2|^2 = 50 - 0$
$|z_1 - z_2|^2 = 50$

The final answer is $\boxed{50}$.