If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$ Problem : 
If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$
Please suggest how to proceed on this question... I got this problem from  [1]: http://www.mathstudy.in/

$$(x+5)^2 +(y-12)^2=14^2$$
$$x^2+y^2-27=24y-10x$$
Now, by Cauchy Schwarz
$$(24y-10x)^2 \leq (24^2+10^2)(x^2+y^2)$$
with equality if and only if $\frac{y}{24}=\frac{x}{-10}$.
Thus
$$(x^2+y^2-27)^2 \leq (24^2+10^2)(x^2+y^2)$$
Let $t := x^2+y^2$. Then
$$(t-27)^2 \leq 676t$$
$$t^2-54t+729 \leq 676 t$$
$$t^2-730t+729 \leq 0$$
or
$$(t-1)(t-729) \leq 0$$
Hence
$$1 \leq t \leq 729$$
This proves that
$$1 \leq x^2+y^2 \leq 729 $$
Note that to get $x^2+y^2=1$ or $x^2+y^2=729$ we must have
$$ \frac{y}{24}=\frac{x}{-10} \,.$$
