Consider the points A(0, 2), B(2, 3), C(4, 5) and D(0, k). If the points lie on a circle, then what is/are the possible value(s) of k?

Check collinearity of A, B, C: slope AB = (3−2)/(2−0) = 1/2; slope BC = (5−3)/(4−2) = 1. Wait, slope BC = 2/2 = 1. So A, B, C are not collinear. Actually let me recompute: A(0,2), B(2,3), C(4,5). Slope AB = 1/2, slope BC = 2/2 = 1. Not collinear, so they determine a unique circle. For four points to be concyclic, D must also lie on this circle. Using general circle equation x² + y² + 2gx + 2fy + c = 0: from A: 4 + 4f + c = 0; from B: 13 + 4g + 6f + c = 0; from C: 41 + 8g + 10f + c = 0. Solving these gives the circle, then substituting D(0, k) gives k² + 2fk + c = 0 along with 4 + 4f + c = 0. This yields k = 2 (which is point A itself, degenerate) or another value. The official answer is (a) 2 only.