Equation of circle touching three circles, two of which are intersecting Find the equation of the circle which is tangentially touching three given circles: $x^2+y^2=49$, $x^2+(y-3.5)^2=49/4$, and $y^2+(x-3.5)^2=49/4$.
By tangentially i mean, it touches the smaller two circle externally and the larger one internally.
The problem would have been much easier had the three given circles been tangent to each other but the smaller two of them intersect, making finding the radius of the circle in question much difficult for me.
Well, I should write the general equation of the circle and equate sum of radius with the distance between the centre with the three given equations.
I am not getting the correct answer with this approach. 
Please help. 

Say the center of the circle is at $(h,k)$, and it’s radius is $r$. 
From the fact that it is touching the “smaller” circles, the distances between its center and their centers are equal to the sum of the radii of the two chosen circles. 
This gives us the following equations:
$$\Bigl(h-\frac 72\Bigr)^2 + k^2 = \Bigl(r + \frac 72\Bigr)^2…(1)$$
$$h^2 + \Bigl(k-\frac 72\Bigr)^2 = \Bigl(r + \frac 72\Bigr)^2…(2)$$
Solving them simultaneously gives us this relation:
$$h=k…(3)$$
Now, since it touches the “large” circle internally, the distance between their centers is the difference between their radii. 
So, we have 
$$h^2 + k^2 = (7-r)^2…(4)$$
Solving $(2), (3) and (4)$, it is now a trivial task to find the required circles:
$$\left(x-\frac{14}{3\sqrt{2}-1}\right)^2+\left(y-\frac{14}{3\sqrt{2}-1}\right)^2=\left(\frac{7\left(\sqrt{2}-1\right)}{3\sqrt{2}-1}\right)^2$$
and 
$$\left(x+\frac{14}{3\sqrt{2}+1}\right)^2+\left(y+\frac{14}{3\sqrt{2}+1}\right)^2=\left(\frac{7\left(\sqrt{2}+1\right)}{3\sqrt{2}+1}\right)^2$$
