A pair of tangents are drawn from the origin | vedclass

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(C) The equation of the pair of tangents from a point $(x_1, y_1)$ to the circle $S = 0$ is given by $SS_1 = T^2$.Here,$S = x^2 + y^2 + 20(x + y) + 20

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$2x^2 + 2y^2 + 5xy = 0$

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(C) The equation of the pair of tangents from a point $(x_1, y_1)$ to the circle $S = 0$ is given by $SS_1 = T^2$.Here,$S = x^2 + y^2 + 20(x + y) + 20$ and the point is $(0, 0)$.$S_1$ is the value of $S$ at $(0, 0)$,so $S_1 = 0^2 + 0^2 + 20(0 + 0) + 20 = 20$.$T$ is the equation of the tangent at $(0, 0)$,given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.Substituting the values,$T = 0 + 0 + 10(x + 0) + 10(y + 0) + 20 = 10(x + y + 2)$.Now,$SS_1 = T^2$ becomes $20(x^2 + y^2 + 20x + 20y + 20) = [10(x + y + 2)]^2$.$20(x^2 + y^2 + 20x + 20y + 20) = 100(x^2 + y^2 + 4 + 2xy + 4x + 4y)$.Dividing by $20$,$x^2 + y^2 + 20x + 20y + 20 = 5(x^2 + y^2 + 2xy + 4x + 4y + 4)$.$x^2 + y^2 + 20x + 20y + 20 = 5x^2 + 5y^2 + 10xy + 20x + 20y + 20$.$4x^2 + 4y^2 + 10xy = 0$.Dividing by $2$,we get $2x^2 + 2y^2 + 5xy = 0$.

$A$ pair of tangents are drawn from the origin to the circle $x^2 + y^2 + 20(x + y) + 20 = 0$. The equation of the pair of tangents is

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