Tangents are drawn from the point A(-2, 1) to | vedclass

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(C) Let the circle be $S: x^2 + y^2 - 4x - 6y + 8 = 0$. The center of the circle is $C(2, 3)$ and the radius is $r = \sqrt{2^2 + 3^2 - 8} = \sqrt{4 +

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$x^2 + y^2 - 4y - 1 = 0$

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(C) Let the circle be $S: x^2 + y^2 - 4x - 6y + 8 = 0$. The center of the circle is $C(2, 3)$ and the radius is $r = \sqrt{2^2 + 3^2 - 8} = \sqrt{4 + 9 - 8} = \sqrt{5}$.Since $AP$ and $AQ$ are tangents from $A(-2, 1)$ to the circle,the angles $\angle APC$ and $\angle AQC$ are $90^{\circ}$.Thus,the points $P$ and $Q$ lie on a circle with diameter $AC$.The midpoint of $AC$ is $(\frac{-2+2}{2}, \frac{1+3}{2}) = (0, 2)$.The equation of the circle with diameter $AC$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting the coordinates of $A(-2, 1)$ and $C(2, 3)$:$(x + 2)(x - 2) + (y - 1)(y - 3) = 0$$x^2 - 4 + y^2 - 4y + 3 = 0$$x^2 + y^2 - 4y - 1 = 0$.

Tangents are drawn from the point $A(-2, 1)$ to the circle $x^2 + y^2 - 4x - 6y + 8 = 0$ touching it at the points $P$ and $Q$. Find the equation of the circle circumscribing the $\Delta APQ$.

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