The equations of the tangents drawn from the point 0-1 to the circle x2-y2-2 x-4 y-0 areA- 2 x - y -1-0- x -2 y -2-0B- 2 x-y-1-0- x-2 y-2-0C- 2 x-y-1-0- x-2 y-2-0D- 2 x-y-1-0- x-2 y-2-0

The correct option is A 2x y + 1 = 0, x + 2y 2 = 0Required equations are given by SS_1 = T^2 ⇒ (x^2 + y^2 2x + 4y)(1+4)=y 1(x) + 2(y+1)^2 ⇒ 2x^2 2y^2 ...

The equations...

Question

The equations of the tangents drawn from the point (0, 1) to the circle $x^{2} + y^{2} - 2 x + 4 y = 0$ are

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

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

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

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

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Solution

The correct option is A 2x - y + 1 = 0, x + 2y - 2 = 0
Required equations are given by $S S_{1} = T^{2}$
$\Rightarrow (x^{2} + y^{2} - 2 x + 4 y) (1 + 4) = y - 1 (x) + 2 (y + 1)^{2}$
$\Rightarrow 2 x^{2} - 2 y^{2} - 3 x + 4 y + 3 x y - 2 = 0$
$\Rightarrow (2 x - y + 1) (x + 2 y - 2) = 0.$

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The correct option is A 2x - y + 1 = 0, x + 2y - 2 = 0
Required equations are given by $S S_{1} = T^{2}$
$\Rightarrow (x^{2} + y^{2} - 2 x + 4 y) (1 + 4) = y - 1 (x) + 2 (y + 1)^{2}$
$\Rightarrow 2 x^{2} - 2 y^{2} - 3 x + 4 y + 3 x y - 2 = 0$
$\Rightarrow (2 x - y + 1) (x + 2 y - 2) = 0.$