There may be an easier way, but I'd just try to find an equation for the circle, on the cartesian plane.

The green circle I presume has radius 20, so the equation is

$x^2 + y^2 = 20^2$

The red circle has unknown radius and unknown shift, so the equation is:

$(x - h)^2 + (y - h)^2 = c^2$

We know 3 points on the red circle, based on your drawing:
$(0,20)$
$(20,0)$
$(-18 \sin \frac{ \pi}{4}, 18 \sin \frac{ \pi}{4})$

Now with this information you can solve for the equation of the red circle and hence the radius.

distance between centers = x

radius of red circle = 18 + x

there is a triangle with sides 20, x and 18 + x and an angle 135 degrees

You could torture yourself working this all out
or you could just use WolframAlpha
circle through (-20, 0), (-9*sqrt (2), 9*sqrt (2)), (0, 20) - Wolfram|Alpha Results

Idea's solution (post 3) requires only the solution of a linear equation; not much torture here.

Originally Posted by johng
Idea's solution (post 3) requires only the solution of a linear equation; not much torture here.
Can't argue with that. Still torture though when I took a whack at it totally sober using
(-20-h)^2-(0-k)^2=r^2
(-9*2^.5-h)^2+(9*2^.5-k)^2=r^2
(0-h)^2-(20-k)^2=r^2

Moral: Never take on someone else's math troubles sober. You'll end up suffering needlessly.

Originally Posted by jonah
(-20-h)^2-(0-k)^2=r^2
(-9*2^.5-h)^2+(9*2^.5-k)^2=r^2
(0-h)^2-(20-k)^2=r^2
Major screw up. Should be
(-20-h)^2+(0-k)^2=r^2
(-9*2^.5-h)^2+(9*2^.5-k)^2=r^2
(0-h)^2+(20-k)^2=r^2