Solution: (1) Because the discriminant of equation 1 △ = k 2-2k+1= (k-1) 2 ≥ 0.

So the equation 1/2x 2+KX+K- 1/2 = 0 has two real roots;

(2) Let the coordinates of two intersections of quadratic function image and X axis be A(x 1, 0) B(x2, 0), then |x 1-x2|=4.

According to the relationship between the roots and coefficients of a quadratic equation, x1+x2 =-2kx1x2 = 2k-1.

So (x1-x2) 2 =16 (x1+x2) 2-4x1x2 =16, that is, (-2k) 2-4 (2k-/kloc-).

So y = 1/x 2-x-3/2.

(3) Under the condition of (2), C( 1, -2), A (- 1, 0) and B (3 3,0) obviously △ABC is a right triangle, so the circumscribed circle of △ABC takes AB midpoint (1, 0).

Ask if you have any questions, and adopt if you have no questions.