The ratio of two circles is $r_{1}$ and $r_{2}$ . If the two circles touch externally, then the length of their direct common tangent is __________.

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Solution

TMA $C_{1} C_{2} E$
$C_{1} C_{2}^{2} = C_{1} E^{2} + C_{2} E^{2}$
${(r_{1} - r_{2})}^{2} = {(r_{1} + r_{2})}^{2} + C_{2} E^{2}$
$C_{2} E^{2} = {(r_{1} + r_{2})}^{2} - {(r_{1} - r_{2})}^{2}$
$C_{2} E^{2} = 4 r_{1} - r_{2}$
$C_{2} E = 2 \sqrt{r_{1} r_{2}}$
$C_{3} E^{2} = {E F}^{2} + C_{3} E^{2}$
$C_{3} F^{2} = C_{3} E^{2} - {E F}^{2}$
$C_{3} F^{2} = {(r + r_{1})}^{2} - {(r_{1} - r)}^{2}$
$C_{3} F^{2} = 4 r r_{1}$
$C_{3} F = 2 \sqrt{r r_{1}}$
$C_{3} G = 2 \sqrt{r r_{2}}$
$C_{3} G + C_{3} F = C_{2} E$
$\sqrt{r r_{2}} + \sqrt{r r_{1}} = \sqrt{r_{1} r_{2}}$
$\sqrt{r} ({\sqrt{r}}_{2} + {\sqrt{r}}_{1}) = \sqrt{r_{1} r_{2}}$
$r = \frac{r_{1} r_{2}}{{(\sqrt{r_{2}} + \sqrt{r_{1}})}^{2}}$
Hence this is the answer

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