minimum $x^2 + y^2$ on $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1 $ ellipse Given  $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1$.
Then minimum value of $x^2 + y^2 = ?$
P.S. My solution: Suppose that $x = 4\cos{\theta}+12$and $y = 5\sin{\theta}-5$
and expand $x^2 + y^2$ to find minimum value, but stuck in the end.
Thank you for every comment.

Another, maybe not as elegant, way is to solve for $y$, getting $$y=-5\pm\frac54\sqrt{16-\left( x -12\right)^2}$$ and then, since $$y_+(x):=\left(-5+\frac54\sqrt{16-\left( x -12\right)^2}\right)^2\le\left(-5-\frac54\sqrt{16-\left( x -12\right)^2}\right)^2=:y_-(x), $$ find the zero $x_0$ of the derivative of $x^2+(y_+(x))^2$ , which is $$-\frac98 x +\frac{25}{2}\left(3+\frac{x-12}{\sqrt{16-(x-12)^2}}\right) \tag{$\star$}$$ and compute $x_0^2+(y_+(x_0))^2$.
Note that $x_0$ is unique and is approximately $8.345$. The other solution mentioned by Mirko is produced by squaring, which is necessary to solve exactly $(\star)$, ending up with $$\frac{(x-12)^2}{16-(x-12)^2}=\left(\frac{9}{100}x-3\right)^2,$$ whose solutions are indeed the roots of the quartic found by Mirko.
