What is the length of common tangents of the circles $x^{2} + y^{2} = 6 x, x^{2} + y^{2} + 2 x = 0$

\sqrt{3}

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\sqrt{3}, 3 \sqrt{3}

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2 \sqrt{3}

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2 \sqrt{3}, 3 \sqrt{3}

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Solution

The correct option is B $2 \sqrt{3}$
$x^{2} + y^{2} - 6 x = 0$
$C_{1} = (3, 0), r_{1} = 3$
and $x^{2} + y^{2} + 2 x = 0$
$C_{2} = (- 1, 0), r_{2} = 1$
$\Rightarrow r_{1} + r_{2} = 4 = C_{1} C_{2} = 4$
So, this two circles touching each other externally.
As we know, point of intersection of common tangent divides segment of line joining centre in ration of there radius.
So, $h = \frac{3 \times (- 1) - 1 (3)}{3 - 1} = - 3$
and $k = \frac{3 \times (0) - 1 \times (0)}{3 - 1} = 0$
So, $P (- 3, 0)$ from P length of external common tangent from point $(1, 2)$
$= \sqrt{d^{2} - (r_{1} - r_{2})^{2}}$
$= \sqrt{4^{2} - 2^{2}}$
$= \sqrt{12}$
$= 2 \sqrt{3}$

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