Use $u^2+v^2=100$ to rewrite $xy=\frac{v^2-u^2}{4}$ using only $u$.

For a fixed value of $u$, consider how many possible values of $v$ satisfy $u^2+v^2=100$. Determine when there can be only one.

Algebra is the fastest method because the substitutions $u=x-y$ and $v=x+y$ explain why most values produce two ordered pairs. Desmos can verify the endpoint result.

Enter $x^2+y^2=100$ and $25-x^2/2+12x=k$. Let Desmos create a slider for $k$.

Adjust the positive $k$ slider until the second graph has exactly one intersection with the circle. The intersection occurs on the horizontal axis at the rightmost point of the circle, confirming the positive value of $k$.

Let $u=x-y$ and $v=x+y$. Then

Therefore, $-10\leq u\leq 10$.

Because

and $v^2=100-u^2$, it follows that

The second equation becomes

For any value strictly between $-10$ and $10$, the equation $u^2+v^2=100$ gives two values of $v$: one positive and one negative. These produce two different ordered pairs $(x,y)$.

At $u=-10$ or $u=10$, however, $v=0$, so there is only one ordered pair.

The expression $25-\frac{u^2}{2}+12u$ increases on the interval from $-10$ to $10$. For $u_2>u_1$ in this interval,

and the factor in parentheses is at least $2$, so the difference is positive.

Thus, evaluate the endpoints:

Since $k$ is positive, its value is $95$.