Question

How do you find the equation for the circle where A(7,-6),B(9,2) and C(1,4) are points on a circle?

Answer 1

`(x-4)^2+(y+1)^2 = 34`

Explanation:

Notice that `vec(AB) = (2,8)` and `vec(BC) = (-8,2)` are perpendicular to one another and of the same length.

So `ABC` is an isosceles right angled triangle. Together with the point `D(-1, -4)` the points `A,B,C,D` form the vertices of a square with centre at the midpoint of `AC`:

`E = ((7+1)/2, (-6+4)/2) = (4, -1)`

The radius of the circumscribing circle is the distance between the centre `E` and any of `A,B,C,D` say `C`:

`r = sqrt((1-4)^2+(4-(-1))^2) = sqrt(3^2+5^2) = sqrt(34)`

The equation of a circle with centre `(h,k)` and radius `r` may be written:

`(x-h)^2+(y-k)^2 = r^2`

Hence the equation of our circle may be written:

`(x-4)^2+(y-(-1))^2 = 34`

or slightly easier to read as:

`(x-4)^2+(y+1)^2 = 34`

graph{((x-4)^2+(y+1)^2-34)((x-4)^2+(y+1)^2-0.03)((x-7)^2+(y+6)^2-0.08)((x-9)^2+(y-2)^2-0.08)((x+1)^2+(y+4)^2-0.08)((x-1)^2+(y-4)^2-0.08) = 0 [-14.67, 25.33, -10.44, 9.56]}