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(A) The equation of the circle is $x^2+y^2+6x-6y+2=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-3, c=2$. The center is $C(-3, 3)$ and the

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(A) The equation of the circle is $x^2+y^2+6x-6y+2=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-3, c=2$. The center is $C(-3, 3)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+9-2} = \sqrt{16} = 4$. Let $	heta$ be the angle between the tangents. Given $	heta = 2 \operatorname{Tan}^{-1}\left(\frac{4}{3}ight)$,so $\frac{	heta}{2} = \operatorname{Tan}^{-1}\left(\frac{4}{3}ight)$,which implies $	an\left(\frac{	heta}{2}ight) = \frac{4}{3}$. In the right-angled triangle formed by the center,the point $P$,and the point of tangency,we have $\sin\left(\frac{	heta}{2}ight) = \frac{r}{CP}$. Since $	an\left(\frac{	heta}{2}ight) = \frac{4}{3}$,we have $\sin\left(\frac{	heta}{2}ight) = \frac{4}{5}$. Thus,$\frac{4}{CP} = \frac{4}{5}$,which gives $CP = 5$. The distance $CP^2 = (k - (-3))^2 + (6k - 3)^2 = 5^2 = 25$. $(k+3)^2 + (6k-3)^2 = 25$. $k^2 + 6k + 9 + 36k^2 - 36k + 9 = 25$. $37k^2 - 30k + 18 = 25$. $37k^2 - 30k - 7 = 0$. $(37k + 7)(k - 1) = 0$. Since $k$ must be an integer,$k = 1$.

The angle between the tangents drawn from the point $P(k, 6k)$ to the circle $x^2+y^2+6x-6y+2=0$ is $2 \operatorname{Tan}^{-1}\left(\frac{4}{3}\right)$. If the coordinates of $P$ are integers,then $k=$

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