For which values of $x,y$ does $x^2+y^2$ take the minimum? 
Find the values of $x,y$ for which $x^2+y^2$ takes the minimum value where $(x+5)^2+(y-12)^2=14^2$.

Trial:$\begin{align} \ (x+5)^2+(y-12)^2=14^2 \\ \implies x^2+y^2+10x-24y=27 \end{align}$.
Here I am stuck. Please help.
 A: What about Lagrange's Multipliers? Let
$$f(x,y)=x^2+y^2\,\,,\,\,g(x,y)=(x+5)^2+(y-12)^2-14^2\Longrightarrow$$
$$H(x,y,\lambda)=f+\lambda g=x^2+\lambda(x+5)^2+y^2+\lambda(y-12)^2-14^2\Longrightarrow$$
$$H'_x=2\left((x+\lambda(x+5)\right)=0\Longrightarrow x=-\frac{5\lambda}{1+\lambda}$$
$$H'_y=2\left(y+\lambda(y-12)\right)=0\Longrightarrow y=\frac{12\lambda}{1+\lambda}$$
$$H'_\lambda=(x+5)^2+(y-12)^2-14^2\Longrightarrow H'_\lambda=g(x,y)=0$$
Substituting the values for $\,x,y\,$ in the last equation above, we get:
$$\left(-\frac{5\lambda}{1+\lambda}+5\right)^2+\left(\frac{12\lambda}{1+\lambda}\right)^2=14^2\Longrightarrow (25+144)\frac{1}{(1+\lambda)^2}=14^2\Longrightarrow$$
$$(1+\lambda)^2=\frac{13^2}{14^2}\Longrightarrow 1+\lambda =\pm\frac{13}{14}$$
Try now to take it from here.
A: Think of it geometrically. You have a point $P$ inside a circle, and you want to find the point on the circle closest to $P$. So, you draw a ray from the center of the circle to/through $P$, and see where that ray intersects the circle. 
A: We can also look at this geometrically:
Letting $r^2 = x^2 + y^2$, we see that each value of $x^2 + y^2$ corresponds to a circle of a different radius, centered at the origin. If we construct the smallest such circle which intersects a circle of radius $14$ centered at $(-5,12)$, the point of intersection will give us our $(x,y)$. A little thought will convince you that the smaller circle will be tangent to the larger circle at this point, so that there is only one solution. But since those circles are tangent, that point will lie on the line connecting their centers (where both circles intersect that line.)
