<s>[INST] 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. 
 [/INST] 
Let $a$ be the radius of the inscribed circle and apply the cosine rule to the triangle formed by the centers of the large circle, one of the small circles and the inscribed circle,
$$(a+3.5)^2=(7-a)^2+3.5^2-2\cdot 3.5 \cdot (7-a)\cos 135^\circ$$
which yields 
$$a= \frac{7(\sqrt{2}+1)}{3\sqrt{2}+1}$$
Therefore, the equation of the inscribed circle is
$$\left( x+ \frac{7-a}{\sqrt{2}} \right)^2
+\left( y+ \frac{7-a}{\sqrt{2}} \right)^2=a^2$$
Similarly, let $b$ be the radius of the smaller inscribed circle. With the cosine rule,
$$(b+3.5)^2=(7-b)^2+3.5^2-2\cdot 3.5 \cdot (7-b)\cos 45^\circ$$
which yields 
$$b= \frac{7(\sqrt{2}-1)}{3\sqrt{2}-1}$$
Therefore, the equation of the smaller inscribed circle is
$$\left( x- \frac{7-b}{\sqrt{2}} \right)^2
+\left( y- \frac{7-b}{\sqrt{2}} \right)^2=b^2$$
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