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.
The solution process is shown in the above four screenshots.
The four questions you uploaded all belong to the combination of factorization, squar