Two circles of radii r₁ & r₂ (r₁ > r₂) touch each other externally then the radius of the circle which touches both of them externally & also their direct common tangent is

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$\frac{r_{1} r_{2}}{2}$

$\frac{r_{1} r_{2}}{{(\sqrt{r_{1}} + \sqrt{r_{2}})}^{2}}$

$r_{1} - r_{2}$

$\sqrt{r_{1} r_{2}}$

answer is A.

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Detailed Solution

$\begin{array}{l} AB = \sqrt{{(r_{1} + r_{2})}^{2} - {(r_{1} - r_{2})}^{2}} = 2 \sqrt{r_{1} r_{2}} \\ A B = A C + C B \\ A B = \sqrt{{(r_{1} + r)}^{2} - {(r_{1} - r)}^{2}} + \sqrt{{(r_{2} + r)}^{2} + {(r_{1} - r)}^{2}} \\ 2 \sqrt{r_{1} r_{2}} = 2 \sqrt{r_{1} r} + 2 \sqrt{r r_{2}} \\ \sqrt{r} = \frac{\sqrt{r_{1} r_{2}}}{\sqrt{r_{1}} + \sqrt{r_{2}}} ∴ r = \frac{r_{1} r_{2}}{{(\sqrt{r_{1}} + \sqrt{r_{2}})}^{2}} \end{array}$