A circle C1 of radius b touches the circle x2 + y2 = a2 externally and has its centre on the positive x-axis. Another circle C2 of radius c touches the circle C1 externally and has its centre on the positive x-axis. If a < b < c, then the three circles have a common tangent if a, b, c are in
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A.P.
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G.P.
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H.P.
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None of these
For three circles with radii a, b, c to have a common tangent, their radii must be in geometric progression (b^2 = ac).
The centers of all three circles lie on the x-axis, so the common tangent must be perpendicular to the x-axis to touch all three. Let the common vertical tangent intersect the x-axis at (d, 0). The tangent must lie to the left of the first circle, meaning d equals -a, and it must lie to the right of circle C1 and C2. The distance from the center of C1 at (a+b, 0) to the tangent line x = -a is the radius b, so a+b - (-a) = b, which implies 2a = 0. This indicates the tangent must be the y-axis (x = 0) intersecting the origin, requiring d = 0 and verifying the first circle has radius a and is centered at the origin. For the second circle to touch the first externally and have its center on the positive x-axis, its center is at (a+b, 0). For a vertical line x = 0 to be tangent to the second circle, the distance from (a+b, 0) to the y-axis must equal the radius b, making a+b = b, which is impossible for a > 0. The problem requires a common external tangent above the x-axis. Using the external homothety center at the origin, the tangent line from the origin to the first circle has length a and slope m, so the line is y = mx. If this same line is tangent to the second circle centered at (a+b, 0) with radius b, the distance from (a+b, 0) to the line mx - y = 0 is b. Using the point-to-line distance formula, we get m(a+b) divided by the square root of m squared plus 1 equals b. Since the line is tangent to the first circle, the distance from (a, 0) to mx - y = 0 equals a, which simplifies to m squared plus 1 equals m squared. This gives m squared equals m squared plus 1 minus 1, so the tangent slope condition implies a over the square root of 1 + m squared equals 1. Working through the distance equality for both circles results in the proportion a/b = b/c, meaning b squared equals ac. This is the defining property of a geometric progression.