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(B) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.Here,$S = x^2 + y^2 - 2x + 4y + 3 = 0$

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$7x^2 + 23y^2 + 30xy + 66x + 50y - 73 = 0$

Vedclass · Answer

(B) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.Here,$S = x^2 + y^2 - 2x + 4y + 3 = 0$.For the point $(6, -5)$,$S_1 = (6)^2 + (-5)^2 - 2(6) + 4(-5) + 3 = 36 + 25 - 12 - 20 + 3 = 32$.The equation of the tangent $T$ at $(x, y)$ is $xx_1 + yy_1 - (x + x_1) + 2(y + y_1) + 3 = 0$.Substituting $(6, -5)$,we get $T = 6x - 5y - (x + 6) + 2(y - 5) + 3 = 5x - 3y - 13$.Now,$SS_1 = T^2$ becomes:$32(x^2 + y^2 - 2x + 4y + 3) = (5x - 3y - 13)^2$.Expanding both sides:$32x^2 + 32y^2 - 64x + 128y + 96 = 25x^2 + 9y^2 - 30xy - 130x + 78y + 169$.Rearranging the terms:$7x^2 + 23y^2 + 30xy + 66x + 50y - 73 = 0$.

The equation of the pair of tangents to the circle $x^2 + y^2 - 2x + 4y + 3 = 0$ from the point $(6, -5)$ is:

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