Points E(-2, -1) and F(3, 2) are the ends of the diameter of a circle....
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: D. \(x^{2} + y^{2} - x - y - 8 = 0\)
Explanation
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.
\(= (x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{34}{4}\)
\(x^{2} - x + \frac{1}{4} + y^{2} - y + \frac{1}{4} = \frac{17}{2}\)
= \(x^{2} - y^{2} - x - y - 8 = 0\)