Points A (-3; 0) and B (3; 6) are given. Equate a circle whose diameter is AB.

As you know, the equation of a circle centered at the point M (x0; y0) and radius R, has the form: (x – x0) ² + (y – y0) ² = R². In order to fulfill the requirement of the task, we need to determine the coordinates of the center of the circle, that is, the point M (x0; y0) and the radius R.
According to the specification, the segment AB is the diameter. This means that, firstly, the center of the circle is in the middle of this segment, and secondly, half the length of the segment is equal to the radius R. Therefore, we have: x0 = (-3 + 3) / 2 = 0/2 = 0 and y0 = (0 + 6) / 2 = 6/2 = 3.
Using the formula for determining the length of a segment with known coordinates of its ends, we get: AB = √ ((3 – (-3)) ² + (6 – 0) ²) = √ (36 + 36) = √ (36 * 2) = √ (36) * √ (2) = 6 * √ (2). Therefore, R = 6 * √ (2) / 2 = 3 * √ (2).
Thus, the required equation has the form: (x – 0) ² + (y – 3) ² = (3 * √ (2)) ² or x² + (y – 3) ² = 18.
Answer: x² + (y – 3) ² = 18.