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

Apr 11, 2016

The equation of the circle may be written:

${\left(x - 3\right)}^{2} + {\left(y - 4\right)}^{2} = {5}^{2}$

#### 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 $\left(0 , 0\right)$ and $\left(0 , 8\right)$ we find that the centre lies on the line $y = 4$.

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

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

So the equation of the circle may be written:

${\left(x - 3\right)}^{2} + {\left(y - 4\right)}^{2} = {5}^{2}$

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]}