If the tangent drawn at the point P on the ci | vedclass

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(A) The equation of the circle is $x^2+y^2+6x+6y-2=0$. The center $O$ is $(-3, -3)$ and the radius $r = \sqrt{(-3)^2+(-3)^2-(-2)} = \sqrt{9+9+2} = \sq

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$5$

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(A) The equation of the circle is $x^2+y^2+6x+6y-2=0$. The center $O$ is $(-3, -3)$ and the radius $r = \sqrt{(-3)^2+(-3)^2-(-2)} = \sqrt{9+9+2} = \sqrt{20}$.Since point $Q$ lies on the $Y$-axis,its $x$-coordinate is $0$. Substituting $x=0$ into the line equation $5x-2y+6=0$,we get $-2y+6=0$,so $y=3$. Thus,$Q$ is $(0, 3)$.The length of the tangent $PQ$ from point $Q(0, 3)$ to the circle is given by $\sqrt{S_1}$,where $S_1 = x_1^2+y_1^2+6x_1+6y_1-2$.$PQ = \sqrt{0^2+3^2+6(0)+6(3)-2} = \sqrt{0+9+0+18-2} = \sqrt{25} = 5$.

If the tangent drawn 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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