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(B) Let $S = x^2 + y^2 - 2x + 4y = 0$. For the point $(x_1, y_1) = (0, 1)$,we have $S_1 = 0^2 + 1^2 - 2(0) + 4(1) = 5$.The equation of the tangent $T$

Vedclass · Accepted Answer

$4x^2 - 4y^2 + 6xy - 6x + 8y - 4 = 0$

Vedclass · Answer

(B) Let $S = x^2 + y^2 - 2x + 4y = 0$. For the point $(x_1, y_1) = (0, 1)$,we have $S_1 = 0^2 + 1^2 - 2(0) + 4(1) = 5$.The equation of the tangent $T$ at $(x_1, y_1)$ is given by $x x_1 + y y_1 - (x + x_1) + 2(y + y_1) = 0$.Substituting $(0, 1)$,we get $T = x(0) + y(1) - (x + 0) + 2(y + 1) = -x + 3y + 2$.The equation of the pair of tangents is given by $SS_1 = T^2$.$5(x^2 + y^2 - 2x + 4y) = (-x + 3y + 2)^2$.$5x^2 + 5y^2 - 10x + 20y = x^2 + 9y^2 + 4 - 6xy - 4x + 12y$.Rearranging the terms,we get $4x^2 - 4y^2 + 6xy - 6x + 8y - 4 = 0$.

The equation of the pair of tangents drawn from the point $(0, 1)$ to the circle $x^2 + y^2 - 2x + 4y = 0$ is . . . . . .

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