Find the length of direct common tangent for circles $x^{2} + y^{2} - 20 x + 64 = 0$ and $x^{2} + y^{2} + 30 x + 144 = 0$

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Solution

The correct option is C

Given circles

$x^{2} + y^{2} - 20 x + 64 = 0$ - - - - - - (1)

$x^{2} + y^{2} + 30 x + 144 = 0$ - - - - - - (2)

Let the circle be $c_{1} & c_{2}$ and radii $r_{1} & r_{2}$ of circle (1) and (2) respectively.

$c_{1} (10, 0)$ $c_{2} (- 15, 0)$

$c_{1} c_{2} = \sqrt{(25)^{2} + 0} = 25$

$r_{1} = \sqrt{g^{2} + f^{2} - c} = \sqrt{100 + 0 + 64} = 6$

$r_{2} = \sqrt{g^{2} + f^{2} - c} = \sqrt{225 + 0 - 144} = 9$

$c_{1} c_{2} > r_{1} + r_{2}$

Both the circle neither touch nor intersect to each other.

Length of direct common tangent

$= \sqrt{d^{2} - (r_{1} - r_{2})^{2}}$

$= \sqrt{(25)^{2} - (6 - 9)^{2}}$

$= \sqrt{625 - 9} = \sqrt{616} = 2 \sqrt{154}$

Length of direct common tangent = $2 \sqrt{154}$ units

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Find the length of direct common tangent for circles $x^{2} + y^{2} - 20 x + 64 = 0$ and $x^{2} + y^{2} + 30 x + 144 = 0$

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Find the length of direct common tangent for circles x2 + y2 − 20x + 64 = 0 and x2 + y2 + 30x + 144 = 0

Find the length of direct common tangent for circles $x^{2} + y^{2} - 20 x + 64 = 0$ and $x^{2} + y^{2} + 30 x + 144 = 0$