The equations of the tangents drawn from the | vedclass

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Question

(A) Let the circle be $S = x^2 + y^2 - 2x + 4y = 0$. For the point $(x_1, y_1) = (0, 1)$,the power of the point is $S_1 = 0^2 + 1^2 - 2(0) + 4(1) = 5$

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

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(A) Let the circle be $S = x^2 + y^2 - 2x + 4y = 0$. For the point $(x_1, y_1) = (0, 1)$,the power of the point is $S_1 = 0^2 + 1^2 - 2(0) + 4(1) = 5$. 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 $SS_1 = T^2$. $5(x^2 + y^2 - 2x + 4y) = (-x + 3y + 2)^2$. $5x^2 + 5y^2 - 10x + 20y = x^2 + 9y^2 + 4 - 6xy - 4x + 12y$. $4x^2 - 4y^2 + 6xy - 6x + 8y - 4 = 0$. Factoring this expression,we get $(2x - y + 1)(x + 2y - 2) = 0$. Thus,the equations of the tangents are $2x - y + 1 = 0$ and $x + 2y - 2 = 0$.

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

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