The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from each other. Let P be the mid point of the line segment joining the centres of C1 and C2 and C be a circle touching circles C1 and C2 externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 and C, then the radius of the circle C is

O,P=3, O,A=1
AP = √(O,P² - O,A²) = √(3² - 1²) = √8 = 2√2
OP² - r² + PA² = r² + (2√2)²
OP² = 8 + r² (1)
OO₂² = OP² + PO₂²
(r+1)² = 8 + r² + 3²
2r + 1 = r² + 17
r² - 2r + 16 = 0
This quadratic equation has no real roots. There must be a mistake in the problem statement or the given solution. Let's reconsider the geometry. Let r be the radius of circle C. The distance between the centers of C1 and C2 is 6. The distance between the centers of C1 and C is r+1, and the distance between the centers of C2 and C is r+1. Let the center of C1 be O1, the center of C2 be O2, and the center of C be O. Then O1O2 = 6, O1O = r+1, O2O = r+1. Let P be the midpoint of O1O2. Then OP = 3. In the right triangle O1PO, we have O1P² + OP² = O1O². So 3² + x² = (r+1)², where x is the distance from P to the point where the common tangent touches circle C1. Similarly, in the right triangle O2PO, we have O2P² + OP² = O2O². So 3² + y² = (r+1)², where y is the distance from P to the point where the common tangent touches circle C2. Since the tangent passes through P, we must have x = y. By Pythagorean theorem, we have (r+1)² = 3² + x² = 9 + x² and x² = (r+1)² - 9. The distance between the centers of C1 and C2 is 6. Thus, the distance between the centers of C1 and C is r+1, and the distance between the centers of C2 and C is r+1. The distance between P and O is 3 - x. By the Pythagorean theorem in triangle O1PO, we have (r+1)² = 3² + x². Similarly, (r+1)² = 3² + y². Since x = y, we have x = 2√2. Then (r+1)² = 9 + 8 = 17, so r+1 = √17, and r = √17 - 1. This is a more geometrically sound solution.