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/
 A: Notice that
$$(x+5)^2+(y-12)^2=14^{2}$$
represents a circle whose center is $(-5,12)$ and its radius is $14$.
On the other hand, $x^2+y^2$ represents the square of the distance between the origin and the point $(x,y)$.
Hence, what you want is the point on the circle which is the closest to the origin. Do you think you can get what you want?
A: Since you are presumably not allowed to use (or haven't yet seen) the tools of calculus, you'll need to apply the useful "geometric" fact that the shortest distance to a curve from a point not lying on that curve is along the perpendicular line from the point to the curve.  So you need a perpendicular line from the origin to the circumference of the circle.  
Consider that any radius is perpendicular to the circumference at the point where it meets the circle.  The radius from the center of the circle through the origin will thus supply the perpendicular line you seek.  Find the equation of the line containing that radius and then the (appropriate, since there are two solutions) coordinates of the line's intersection point with the circle.  From there, you can find the square of the distance of that point from the origin.
(The coordinates are rational numbers, so the result isn't "too ugly". EDIT -- In fact, the answer to the question is surprisingly simple.)

A: As mathlove has already identified   the curve to be circle $\displaystyle (x+5)^2+(y-12)^2=14^2$
But I'm not sure how to finish  from where he has left of without calculus. 
Here is one of the ways:
Using Parametric equation, any point $P(x,y)$ on the circle can be represented as $\displaystyle (14\cos\phi-5, 14\sin\phi+12)$ where $0\le \phi<2\pi$
So, $\displaystyle  x^2+y^2=(14\cos\phi-5)^2+(14\sin\phi+12)^2$
$\displaystyle=14^2+5^2+12^2+28(12\sin\phi-5\cos\phi)$
Now  can you derive $$-\sqrt{a^2+b^2}\le a\sin\psi-b\cos\psi\le\sqrt{a^2+b^2}?$$
and complete the answer?
A: $$(x+5)^2 +(y-12)^2=14^2$$
$$x^2+y^2-27=24y-10x$$
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
$$ \frac{y}{24}=\frac{x}{-10} \,.$$
