Points E(-2, -1) and F(3, 2) are the ends of the diameter of a circle. Find the equation of the circle.

A.

\(x^{2} + y^{2} - 5x + 3 = 0\)

B.

\(x^{2} + y^{2} - 2x - 6y - 13 = 0\)

C.

\(x^{2} + y^{2} - x + 5y - 6 = 0\)

D.

\(x^{2} + y^{2} - x - y - 8 = 0\)

Correct answer is D

Given the endpoints of the diameter |EF|, the midpoint is the centre of the circle

= \((\frac{-2 + 3}{2} , \frac{-1 + 2}{2}) = (\frac{1}{2} , \frac{1}{2})\)

The radius is the distance from the centre to any point on the circle. Using \((\frac{1}{2}, \frac{1}{2})\) and \((3, 2)\);

\(r^{2} = (3 - \frac{1}{2})^{2} + (2 - \frac{1}{2})^{2} = \frac{25}{4} + \frac{9}{4}\)

\(r^{2} = \frac{34}{4}\)

The equation of a circle is given as:

\((x - a)^{2} + (y - b)^{2} = r^{2}\), (a, b) as the centre of the circle.

\(= (x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{34}{4}\)

\(x^{2} - x + \frac{1}{4} + y^{2} - y + \frac{1}{4} = \frac{17}{2}\)

= \(x^{2} - y^{2} - x - y - 8 = 0\)