The equations of the tangents drawn from the | vedclass

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(A) The equation of the circle is $S: x^2 + y^2 - 2x + 4y = 0$. Given point is $P(x_1, y_1) = (0, 1)$. The equation of the pair of tangents from $(x_1

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

Vedclass · Answer

(A) The equation of the circle is $S: x^2 + y^2 - 2x + 4y = 0$. Given point is $P(x_1, y_1) = (0, 1)$. The equation of the pair of tangents from $(x_1, y_1)$ to the circle $S=0$ is given by $SS_1 = T^2$. Here,$S = x^2 + y^2 - 2x + 4y$,$S_1 = 0^2 + 1^2 - 2(0) + 4(1) = 5$. $T = xx_1 + yy_1 - (x + x_1) + 2(y + y_1) = x(0) + y(1) - (x + 0) + 2(y + 1) = y - x + 2y + 2 = -x + 3y + 2$. Substituting these into $SS_1 = T^2$: $5(x^2 + y^2 - 2x + 4y) = (-x + 3y + 2)^2$. $5x^2 + 5y^2 - 10x + 20y = x^2 + 9y^2 + 4 - 6xy - 4x + 12y$. $4x^2 - 4y^2 + 6xy - 6x + 8y - 4 = 0$. Dividing by $2$: $2x^2 - 2y^2 + 3xy - 3x + 4y - 2 = 0$. Factoring the expression: $(2x - y + 1)(x + 2y - 2) = 0$. 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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