Multiple choice

Two circles with radii '$a$' and '$b$' respectively touch each other externally. Let '$c$' be the radius of a circle that touches these two circles as well as a common tangent to the two circles. Then

  1. $-\dfrac {1}{\sqrt a}-\dfrac {1}{\sqrt b}=\dfrac {1}{\sqrt c}$
  2. $\dfrac {1}{\sqrt a}-\dfrac {1}{\sqrt b}=-\dfrac {1}{\sqrt c}$
  3. $\dfrac {1}{\sqrt a}+\dfrac {1}{\sqrt b}=\dfrac {1}{\sqrt c}$
  4. $\dfrac {1}{\sqrt a}-\dfrac {1}{\sqrt b}=\dfrac {1}{\sqrt c}$
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C Correct answer
Explanation

For two externally touching circles of radii a and b, a third circle of radius c touching both and their common tangent satisfies the Descartes circle theorem variant for a line (which acts as a circle of infinite radius). The relationship is 1/sqrt(c) = 1/sqrt(a) + 1/sqrt(b).

AI explanation

Let the centers of the circles with radii a and b be at coordinates 0 and the square root of ab, respectively. By creating right triangles from the centers of these circles to the center of the third circle with radius c, you can set up the geometric equations for their horizontal and vertical distances. Solving the system of equations for the tangency points eliminates the variables and results in the formula 1 divided by the square root of a plus 1 divided by the square root of b equals 1 divided by the square root of c.

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