If the tangent at the point P on the circle x | vedclass

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(C) Let the point $P$ be $(x_1, y_1)$. The equation of the tangent at $P$ to the circle $x^2 + y^2 + 6x + 6y - 2 = 0$ is given by $x x_1 + y y_1 + 3(x

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$5$

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(C) Let the point $P$ be $(x_1, y_1)$. The equation of the tangent at $P$ to the circle $x^2 + y^2 + 6x + 6y - 2 = 0$ is given by $x x_1 + y y_1 + 3(x + x_1) + 3(y + y_1) - 2 = 0$. Since the point $Q$ lies on the $y$-axis,its $x$-coordinate is $0$. Given that $Q$ lies on the line $5x - 2y + 6 = 0$,substituting $x = 0$ gives $5(0) - 2y + 6 = 0$,which implies $y = 3$. Thus,$Q = (0, 3)$. Since $Q$ lies on the tangent,it must satisfy the tangent equation: $0(x_1) + 3(y_1) + 3(0 + x_1) + 3(3 + y_1) - 2 = 0$. Simplifying this,we get $3y_1 + 3x_1 + 9 + 3y_1 - 2 = 0$,or $3x_1 + 6y_1 + 7 = 0$. The length $PQ$ is given by $\sqrt{(x_1 - 0)^2 + (y_1 - 3)^2} = \sqrt{x_1^2 + y_1^2 - 6y_1 + 9}$. Since $P$ lies on the circle,$x_1^2 + y_1^2 = 2 - 6x_1 - 6y_1$. Substituting this into the expression for $PQ^2$: $PQ^2 = (2 - 6x_1 - 6y_1) - 6y_1 + 9 = 11 - 6x_1 - 12y_1 = 11 - 2(3x_1 + 6y_1)$. Substituting $3x_1 + 6y_1 = -7$,we get $PQ^2 = 11 - 2(-7) = 11 + 14 = 25$. Therefore,$PQ = 5$.

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

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