The circles of radii a and b touch externally. AB is a common tangent. The radius of the circle which touches the line AB and the two circles externally is
- $\sqrt { a b }$
- $\frac { a + b } { 2 }$
- $\frac { 2 a b } { a + b }$
- $\frac { a b } { ( \sqrt { a } + \sqrt { b } ) ^ { 2 } }$
For two circles of radii a and b touching externally, the radius r of a circle touching both and their common tangent is given by the formula 1/sqrt(r) = 1/sqrt(a) + 1/sqrt(b). Solving for r: 1/sqrt(r) = (sqrt(b) + sqrt(a)) / sqrt(ab), so sqrt(r) = sqrt(ab) / (sqrt(a) + sqrt(b)). Squaring both sides gives r = ab / (sqrt(a) + sqrt(b))^2.
Let the radii be a, b, and r, with centers at heights a, b, and r above the x-axis, respectively. The distance between the centers of the first two circles is a + b, which gives the horizontal distance as sqrt((a+b)^2 - (a-b)^2) = 2sqrt(ab). Applying the Pythagorean theorem to the right triangles formed by the centers of the new circle (radius r) and the original circles gives the horizontal distances from the new circle's center as 2sqrt(ar) and 2sqrt(br). Therefore, 2sqrt(ab) = 2sqrt(ar) + 2sqrt(br), which simplifies to sqrt(ab) = sqrt(r)(sqrt(a) + sqrt(b)). Solving for r, we get r = ab / (sqrt(a) + sqrt(b))^2. The radius is ab / (sqrt(a) + sqrt(b))^2.