lf the circles $(x - a)^{2} + (y - b)^{2} = r^{2}, (x - b)^{2} + (y - a)^{2} = r^{2}$ have three common tangents, then which of the following relation follows:

(a - b)^{2}

= 2 r^{2}

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(a + b)^{2} = 2 r^{2}

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a^{2} + b^{2} = 2 r^{2}

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a^{2} + b^{2} = r^{2}

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Solution

The correct option is A $(a - b)^{2}$ $= 2 r^{2}$
$S_{1} : (x - a)^{2} + (y - b)^{2} = r^{2}$
$C_{1} \equiv (a, b)$ & $r_{1} = r$
$S_{2} : (x - b)^{2} + (y - a)^{2} = r^{2}$
$C_{2} \equiv (b, a)$ & $r_{2} = r$
Given, $S_{1}$ and $S_{2}$ have there common tangents , then $S_{1}$ and $S_{2}$ touches each other externally
$∴ C_{1} C_{2} = r_{1} + R_{2}$
$\sqrt{2} (a - b) = 2 r$
$\Rightarrow (a - b)^{2} = 2 r^{2}$

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