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.

HINT: use the Lagrange Multiplier Method
$$F(x,y,\lambda)=x^2+y^2+\lambda((x-12)^2/16+(y+5)^2/25-1)$$
solve the system
$$2\,x+\lambda\, \left( x/8-3/2 \right) =0$$
$$2\,y+\lambda\, \left( {\frac {2}{25}}\,y+2/5 \right)=0 $$
$$1/16\, \left( x-12 \right) ^{2}+1/25\, \left( y+5 \right) ^{2}-1=0$$
