If the tangent at a point P on the circle x^2 | vedclass

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Question

(C) Let $P = (x_1, y_1)$. The equation of the tangent at $P$ is given by: $xx_1 + yy_1 + 3(x + x_1) + 3(y + y_1) - 2 = 0$ $x(x_1 + 3) + y(y_1 + 3) + 3

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let $P = (x_1, y_1)$. The equation of the tangent at $P$ is given by: $xx_1 + yy_1 + 3(x + x_1) + 3(y + y_1) - 2 = 0$ $x(x_1 + 3) + y(y_1 + 3) + 3x_1 + 3y_1 - 2 = 0 \dots (i)$ Since $Q$ lies on the $y$-axis,its $x$-coordinate is $0$. Given $Q$ is the intersection of the tangent and the line $5x - 2y + 6 = 0$,we substitute $x = 0$ into the line equation: $5(0) - 2y + 6 = 0 \implies 2y = 6 \implies y = 3$. So,$Q = (0, 3)$. Since $Q$ lies on the tangent $(i)$,we substitute $(0, 3)$ into $(i)$: $0(x_1 + 3) + 3(y_1 + 3) + 3x_1 + 3y_1 - 2 = 0$ $3y_1 + 9 + 3x_1 + 3y_1 - 2 = 0 \implies 3x_1 + 6y_1 + 7 = 0$. The length $PQ^2 = (x_1 - 0)^2 + (y_1 - 3)^2 = x_1^2 + y_1^2 - 6y_1 + 9$. Since $P(x_1, y_1)$ is on the circle,$x_1^2 + y_1^2 = 2 - 6x_1 - 6y_1$. Substituting this into the expression for $PQ^2$: $PQ^2 = (2 - 6x_1 - 6y_1) - 6y_1 + 9 = 11 - 6x_1 - 12y_1 = 11 - 2(3x_1 + 6y_1)$. Using $3x_1 + 6y_1 = -7$: $PQ^2 = 11 - 2(-7) = 11 + 14 = 25$. Therefore,$PQ = 5$.

If the tangent at a point $P$ on the circle $x^2 + y^2 + 6x + 6y = 2$ meets the line $5x - 2y + 6 = 0$ at a point $Q$ on the $y$-axis,then the length of $PQ$ is . . . . .

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