1. Mid-point of AB is (2, 1).
2. Slope of line through AB
= \(\frac{2-0}{4-0} = \frac{1}{2}\)
Slope of perpendicular line is – 2
Equation of the perpendicular line to AB is
y – 1 = -2(x – 2) ⇒ 2x + y = 5.
3. The meeting point of perpendicular bisector of AB and AC will be the centre of the circum circle.
The line perpendicular to AC is x = 4
Solving and x = 4
We get y = 5 – 8 = -3 and x = 4
Hence center is (4, -3) and radius is
\(\sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = 5\)
Equation of the circle is
(x – 4)2 + (y + 3)2 = 5.