What are the center and radius of the circle given by x² + y² - 6x + 4y - 12 = 0?

Complete the square in both variables: $(x^2-6x+9)+(y^2+4y+4)=12+9+4$, so $(x-3)^2+(y+2)^2=25$. The standard form $(x-h)^2+(y-k)^2=r^2$ gives center $(3,-2)$ and radius $\sqrt{25}=5$, choice F. Choice G has the signs of the center reversed, $(-3,2)$. Choice H has the right center but reports $r^2=25$ as the radius. Choice J uses the full coefficients, $(6,-4)$, instead of half of them. Check with the graph: the center $(3,-2)$ is shown in the figure, and the point $(8,-2)$, which is $5$ to the right, satisfies the original equation.