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

Vedclass · Accepted Answer

$y=0, (g^2-f^2)y-2gfx=0$

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(C) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S=0$ is given by $T^2=SS_1$.Here,the point is the origin $(0,0)$ and the circle is $S: x^2+y^2+2gx+2fy+g^2=0$.$S_1$ at $(0,0)$ is $0^2+0^2+2g(0)+2f(0)+g^2 = g^2$.The tangent $T$ at $(0,0)$ is $x(0)+y(0)+g(x+0)+f(y+0)+g^2 = gx+fy+g^2$.Substituting into $T^2=SS_1$:$(gx+fy+g^2)^2 = (x^2+y^2+2gx+2fy+g^2)(g^2)$$g^2x^2+f^2y^2+g^4+2gfxy+2g^3x+2g^2fy = g^2x^2+g^2y^2+2g^3x+2g^2fy+g^4$Canceling common terms $g^2x^2, g^4, 2g^3x, 2g^2fy$ from both sides:$f^2y^2+2gfxy = g^2y^2$$y^2(g^2-f^2)-2gfxy = 0$$y[(g^2-f^2)y-2gfx] = 0$Thus,the equations are $y=0$ and $(g^2-f^2)y-2gfx=0$.

The equations of the tangents drawn from the origin to the circle $x^2+y^2+2gx+2fy+g^2=0$ are

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