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

1 Answer
Apr 11, 2016

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 #(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]}