The tangent from the point of intersection of the lines $2 x - 3 y + 1 = 0$ and $3 x - 2 y - 1 = 0$ to the circle $x^{2} + y^{2} + 2 x - 4 y = 0$ is-

x + 2 y = 0, x - 2 y + 1 = 0

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2 x - y - 1 = 0

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y = x, y = 3 x - 2

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2 x + y + 1 = 0

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Solution

The correct option is B $2 x - y - 1 = 0$
Given lines
$2 x - 3 y + 1 = 0$ ....(1)
$3 x - 2 y - 1 = 0$ .....(2)

Solving (1) and (2), we get
$x = 1, y = 1$
So, the point of the intersection of the lines is $(1, 1)$
Given equation of circle is $x^{2} + y^{2} + 2 x - 4 y = 0$
Here, $g = 1, f = - 2, c = 0$
So, eqn of tangent from $(1, 1)$ to the circle is
$x + y + 1 (x + 1) - 2 (y + 1) = 0$
$\Rightarrow 2 x - y - 1 = 0$

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x^{2} + y^{2} - 2 x + 4 y = 0

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