Because the radius can be infinite, A and C are incorrect, and the center of the circle is on (k- 1, 3k) and on the straight line Y=3(X+ 1), so this straight line intersects all the circles, and (k- 1) 2 is obtained by substituting (0,0) into the equation of the circle.

When k = 1, k = 2 and k = 3, all equations are invalid. When k is greater than 3, the equation becomes

k^4-(k- 1)^2+k^4-9k^2=0

(k^2-k+ 1)(k^2+k- 1)+k^2(k^2-9)=0

The left side of the equation is a constant, so the equation has no solution and the circle does not pass through the origin.