Point of Tangency between Two Circles I made sure to look around this site a couple times before asking my question. A lot of people have already asked how to find the point of tangency or intersection between two circles, but none of them could really help me solve the problem.
I'm being asked to find the point of tangency between two circles and all I am given are just the two circles -- no equations of tangent lines, etc. Here's a picture of what I'm working with:

The equation of the green circle is $x^2$ + $y^2$ = 5.
The equation of the orange circle was given as $x^2$ - 12x + $y^2$ + 6y = -25, which I rewrote as ($x-6)^2$ + $(y+3)^2$ = 20.
How do I find the coordinates of the circles' point of tangency?
Thank you in advance!
 A: The equations are:
\begin{gather}
    x^2 + y^2 = 5 \tag{1} \\
    x^2 -12x + y^2 + 6y + 25 = 0 \tag{2}
\end{gather}
If you substitute  (1) into (2), you get
\begin{align}
    -12 x + 6y + 30 &= 0 \implies y=2x-5\tag{3}
\end{align}
Substitute this back into (1):
\begin{align}
    x^2 + (2x-5)^2 &= 5 \\\implies
    x^2 + 4x^2 - 20x + 25 &= 5 \\\implies
    5x^2 - 20x + 20 &= 0 \\\implies
    x^2 -4 x + 4 &= 0 \\\implies
    (x-2)^2 &= 0
\end{align}
Therefore $x=2$.  By (3), $y=-1$.
A: Their centers are $(0,0)$ and $(6, -3)$, and their radii are $\sqrt5$ and $\sqrt {20}$ respectively. Now draw a line segment connecting their centers, and we see that the point of tangency is where this line segment intersects both circles. We see that this line segment has a slope of $-\frac{1}{2}$, so its equation is $y=-\frac{1}{2}x$. Now find where this intersect either circle, and you're good to go. 
A: If two circles are tangent, the point of tangency $P=(p_x,p_y)$ is on the line that goes through their centers $A=(a_x,a_y)$ and $B=(b_x,b_y)$, so it is a linear combination of both:
$$(p_x,p_y) = \frac{dist(A,P)}{dist(A,B)}\ \ (a_x,a_y) + \frac{dist(B,P)}{dist(A,B)}\ \ (b_x,b_y)$$
