Find the length of direct common tangent of circle $(x - 1)^{2} + (y - 2)^{2} = 4$ and $(x - 5)^{2} + (y - 2)^{2} = 1$

\sqrt{17}

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\sqrt{15}

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\sqrt{5}

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\sqrt{7}

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Solution

The correct option is A $\sqrt{17}$
Length of $D C T = \sqrt{{(r_{1} - r_{2})}_{1}^{2} + {(c_{1} c_{2})}_{1}^{2}}$
$\Rightarrow r_{1} = 2, r_{2} = 1$
$\Rightarrow c_{1} c_{2} = \sqrt{{(5 - 1)}^{2} + {(2 - 2)}^{2}} = 4$
$∴ l = \sqrt{1^{2} + 4^{2}}$
$= \sqrt{1 + 16}$
$= \sqrt{17}$
Hence, the answer is $\sqrt{17} .$

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