Four Tangent Circles

Given that this problem states that $A$ is an integer, probably the quickest way to solve this problem is by numerical means. Let $R$ be the unknown radius of the sphere. Then the angles subtended by the radii of the circles at the center of the sphere are

$r1=ArcSin\left( \dfrac { \sqrt { 165 } }{ R } \right)$
$r2=ArcSin\left( \dfrac { \sqrt { 220 } }{ R } \right)$
$r3=ArcSin\left( \dfrac { \sqrt { 297 } }{ R } \right)$
$r4=ArcSin\left( \dfrac { \sqrt { 540 } }{ R } \right)$

so that we have $4$ spherical triangles of sides

$(r1+r2, r2+r3 ,r3+r1)$
$(r1+r2, r2+r4, r4+r1)$
$(r1+r3, r3+r4, r4+r1)$
$(r2+r3, r3+r4, r4+r2)$

The total area of the $4$ spherical triangles of the unit sphere has to be $4\pi$ , and we can use L’Huilier’s formula for the area of a spherical triangle, given the sides

$4ArcTan\sqrt { Tan\left( \frac { 1 }{ 2 } s \right) Tan\left( \frac { 1 }{ 2 } \left( s-a \right) \right) Tan\left( \frac { 1 }{ 2 } (s-b) \right) Tan\left( \frac { 1 }{ 2 } (s-c) \right) }$

where $s$ is the semi-perimeter of the sides of the spherical triangle.

As a starting point in numerical approximation work, we can use the largest radius $r4$ , and it works out that $R$ is indeed numerically the same as $r4$ , and so the answer is $(4)(540)=2160$

Proving that $R$ and $r4$ are identical takes more work.

Consider a tetrahedron of sides

$(1+2,1+3,2+3,3+4,2+6,3+6) = (3,4,5,7,8,9)$

The inscribed circles in the $4$ sides are tangent to where the segments meet, the segments being $1,2,3,6$ as measured from the vertices of the tetrahedron. The radius of an circle inscribed in a triangle, given such distances $x, y, z$ from the vertices to the tangent points is

$r=\sqrt { \dfrac { xyz }{ x+y+z } }$

which will work out to be $\sqrt { 165 } ,\sqrt { 220 } ,\sqrt { 297 } ,\sqrt { 540 }$ , and then it's a matter of finding the radius of the sphere that is tangent to the edges of this tetrahedron, which eventually can be worked out. Not every tetrahedron can have all six of its edges tangent to a sphere, and if such a sphere does exist, then the circles inscribed on all four of its sides share common points of tangency.

Finally, given four arbitrary circles that meet at $6$ points, sharing common tangents, then given the three planes of any three circles, the planes will share a common point in the non-degenerate case, and it's not hard to show that the distances from that common point to the three points of tangency are equal. Then, see above paragraphs describing such a tetrahedron.