I think it looks like a... circle?

Consider a circle $x^2+y^2=c^2$ .

The distance of the point of intersection of this circle with the given circle from the origin gives the value of $x^2+y^2$ , i.e., $c^2$ . It can be easily seen from the figure that the point of intersection of the blue circle with the given circle gives the minimum distance from the origin.

The distance of $(-5,12)$ from the origin is $13$ and the radius of the given circle is $14$ . So, radius of the blue circle, i. e., $c$ is $1$ . So, minimum value of $x^2+y^2$ is $1$ .

$\textbf{Solution outline:}$

The equation can be visualized as a circle centered at $(-5,12)$ and having radius $14$ . So $x^{2}+y^{2}$ is clearly the square of the minimum distance of the circumference of the circle from the origin which is $(14 - 13)^{2} = \boxed{1}$ .

Put x + 5 = 14cos(theta) and y - 12 = 14sin(theta) Thus, x^2 + y^2 = 196cos^2(theta) + 25 - 140cos(theta) + 196sin^2(theta) + 144 + 336sin(theta) = 365 + 336sin(theta) -140cos(theta) Thus, minimum value = 365 - 364 = 1

Get the equation of the line containing the origin (0,0) and the center of the circle (-5,12)

Get the solutions (intersection) of the Circle and the line.

There will be two solution, but use only the coordinate near the origin.

( 5/13 , -12/13) --- this is the solution

x^2 + y^2 = (5/13) ^ 2 + (-12/13) ^ 2 = 1