Multiple choice

Two circles of radii $r_1$ and $r_2(r_1 > r_2)$ touch each other externally. Then the radius of circle which touches both of them externally and also their direct common tangent is?

  1. $\dfrac{r_1r_2}{\left(\sqrt{r_1}+\sqrt{r_2}\right)^2}$
  2. $\sqrt{r_1r_2}$
  3. $\dfrac{r_1+r_2}{2}$
  4. $\dfrac{r_1-r_2}{2}$
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A Correct answer
Explanation

For two circles of radii r1 and r2 touching externally, the radius r of a third circle touching both and their common tangent is given by 1/sqrt(r) = 1/sqrt(r1) + 1/sqrt(r2). Solving for r gives r = r1*r2 / (sqrt(r1) + sqrt(r2))^2.

AI explanation

When a circle of radius r touches two externally tangent circles of radii r1 and r2, all three are mutually tangent to each other, meaning the distances between their centers are r plus r1, r plus r2, and r1 plus r2. We can model this configuration using Descartes' Circle Theorem for four mutually tangent circles, where one tangent line is treated as a circle of curvature 0. Solving the theorem for the curvature k (where k is 1 divided by r) yields the formula r equals r1 times r2 divided by the square of the quantity of the square root of r1 plus the square root of r2.

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