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

##### 1 Answer

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

From the points

So the centre of the circle is at

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]}