Question

How do you write an equation of a circle that passes through (0,0), (0,8), (6,0)?

Answer 1

The equation of the circle may be written:

`(x-3)^2+(y-4)^2 = 5^2`

Explanation:

For any pair of points that lie on the circle, the centre of the circle will lie on a perpendicular line through the midpoint of the line segment joining those two points.

So from the points `(0, 0)` and `(0, 8)` we find that the centre lies on the line `y = 4`.

From the points `(0, 0)` and `(6, 0)` we find that the centre lies on the line `x = 3`.

So the centre of the circle is at `(3, 4)` and it passes throught the origin `(0, 0)` which is at a distance `sqrt(3^2+4^2) = 5` from the centre.

So the equation of the circle may be written:

`(x-3)^2+(y-4)^2 = 5^2`

graph{((x-3)^2+(y-4)^2-5^2)(x^2+y^2-0.04)((x-6)^2+y^2-0.04)(x^2+(y-8)^2-0.04)((x-3)^2+(y-4)^2-0.04)(y-4)(x-3+y*0.0001) = 0 [-8.13, 14.37, -1.665, 9.585]}