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A circle touches both circles (x - 2)² + y² = 1 and (x + 2)² + y² = 1 internally, and also lies along the angle bisectors of the transverse common tangents of these two circles. If the radius of such a circle is expressed as p +/- sqrt(q), find p + q.

1. 8
2. 9
3. 10
4. 11

Correct answer: 9

Solution

The two circles (x+-2)² + y² = 1 each have radius 1 and centres at (+-2, 0). Their transverse common tangents pass through the origin; the angle bisectors of these tangents are the coordinate axes. A circle with centre on the y-axis (0, k) internally tangent to both: distance from (0,k) to (2,0) = sqrt(4+k²) = r-1 (internal tangency). Also by symmetry k can be 0. If k=0: sqrt(4) = r-1 => r = 3. But we need radius = p +/- sqrt(q), suggesting two solutions. For centre on x-axis: centre (h, 0), distance to (2,0) = |h-2| = r-1 and to (-2,0) = |h+2| = r-1, giving h=0 and r=3. For centre on y-axis: centre (0,k), sqrt(4+k²) = r-1 and also sqrt(4+k²) = r-1 (same equation). Need additional condition from the angle bisectors — the circle must also be tangent to the angle bisectors (lines y=x, y=-x)? Distance from (0,k) to y=x line is |k|/sqrt(2) = r. Combined with r = sqrt(4+k²)-1: |k|/sqrt(2) = sqrt(4+k²)-1. Let k² = t: t/2 = (sqrt(4+t)-1)² = 4+t - 2sqrt(4+t) + 1. t/2 = 5+t - 2sqrt(4+t). 2sqrt(4+t) = 5+t/2. 4(4+t) = 25 + 5t + t²/4. 16+4t = 25+5t+t²/4. t²/4 + t + 9 = 0. t² + 4t + 36 = 0. Discriminant negative — no real solution. Re-examine setup.

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