A transverse common tangent to the circles $x^{2} + y^{2} + 4 x + 2 y = 4$ and $x^{2} + y^{2} - 4 x - 2 y + 4 = 0$ is

x = 1

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y = 2

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3 x + 4 y + 5 = 0

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2 x + 3 = 0

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Solution

The correct option is B $y = 2$
p divides $c_{1}$ and $c_{2}$ externally in $3 : 1$
ratio
so, $h = \frac{3 \times 2 - 1 (- 2)}{3 - 1} = \frac{8}{2} = 4$
and $k = \frac{3 \times 1 - 1 (- 1)}{3 - 1} = \frac{4}{2} = 2$
so , $e q^{x}$ of tangent is
$y - 2 = m (x - 4)$
$y = m x + 2 - 4 m$
$1 = | \frac{1 - 2 m + 2 + 4 m}{\sqrt{1 + m^{2}}} |$
$\sqrt{1 + m^{2}} = | 2 m - 1 |$
$1 + m^{2} = 4 m^{2} - 4 m + 1$
$3 m^{2} - 4 m = 0$
$m = 0$ or $m = \frac{4}{3}$
so, $e q^{x}$ of common tangents are
$y = 2$ and $3 y - 4 x = 6 - 4 \times 4 = - 10$

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The common chord of the circle $x^{2} + y^{2} + 4 x + 1 = 0$ and $x^{2} + y^{2} + 6 x + 2 y + 3 = 0$ is

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