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

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(C) The equation of the circle is $S: x^2+y^2+6x+6y-2=0$.Point $Q$ lies on the $Y$-axis and on the line $5x-2y+6=0$.Setting $x=0$ in the line equation

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$5$

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(C) The equation of the circle is $S: x^2+y^2+6x+6y-2=0$.Point $Q$ lies on the $Y$-axis and on the line $5x-2y+6=0$.Setting $x=0$ in the line equation: $5(0)-2y+6=0$ $\Rightarrow -2y=-6$ $\Rightarrow y=3$.So,the coordinates of $Q$ are $(0, 3)$.The length of the tangent from a point $(x_1, y_1)$ to a circle $S=0$ is given by $\sqrt{S_1}$.Substituting $Q(0, 3)$ into the circle equation $S(x, y) = x^2+y^2+6x+6y-2$:$S_1 = 0^2 + 3^2 + 6(0) + 6(3) - 2 = 0 + 9 + 0 + 18 - 2 = 25$.Therefore,the length of the tangent $PQ = \sqrt{S_1} = \sqrt{25} = 5$.

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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