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(C) The equation of the circle is $x^2+y^2-4x+2y-4=0$. The center is $C(2, -1)$ and the radius is $r = \sqrt{2^2 + (-1)^2 - (-4)} = \sqrt{4+1+4} = 3$.

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$x^2+y^2+x-7=0$

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(C) The equation of the circle is $x^2+y^2-4x+2y-4=0$. The center is $C(2, -1)$ and the radius is $r = \sqrt{2^2 + (-1)^2 - (-4)} = \sqrt{4+1+4} = 3$. Let $P = (-3, 1)$. The circle with diameter $PC$ has the equation $(x - (-3))(x - 2) + (y - 1)(y - (-1)) = 0$. This simplifies to $(x+3)(x-2) + (y-1)(y+1) = 0$,which is $x^2+x-6 + y^2-1 = 0$,or $x^2+y^2+x-7=0$. The points of contact $A$ and $B$ lie on this circle because $\angle PAC = 90^\circ$ and $\angle PBC = 90^\circ$. Thus,the circumcircle of $	riangle PAB$ is the circle with diameter $PC$,which is $x^2+y^2+x-7=0$.

If $A$ and $B$ are the points of contact of the tangents drawn from the point $P(-3, 1)$ to the circle $x^2+y^2-4x+2y-4=0$,then the equation of the circumcircle of the triangle $PAB$ is

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