Question

Find the equation of the circle that contains the points (4, 6), (-2, 4), (8, -6)

Answer 1

let's denote 3 points: A(4, 6), B(-2, 4), C(8, -6), then:

$c = |\overline{AB}| = \sqrt{(-2-4)^2 + (4-6)^2} = \sqrt{36 + 4} = \sqrt{40} \newline a = |\overline{BC}| = \sqrt{(8-(-2))^2 + (-6-4)^2} = \sqrt{100+100} = \sqrt{200} \newline b = |\overline{CA}| = \sqrt{(4-8)^2 + (6-(-6))^2} = \sqrt{16+144} = \sqrt{160}$

from here we can conclude that triangle ABC is right triangle with angle of A equals 90 degree, as $a^2 = b^2 + c^2$, then centre of the circumscribed circle is on the middle of the vector BC.

and it is a point O(3, -1)

and the length of the radius is equal: $r = a/2 = \sqrt{50}$

so the equation of the circle will be: $(x-3)^2 + (y+1)^2 = 50$