Question

How do you find the equation of the circle in general form containing the points (-3,-5), (1,3), and (5,-1)?

Answer 1

Use an alternative, slightly more geometric method (not necessarily shorter) to find:

`(x-1/3)^2+(y+5/3)^2 = ((10 sqrt(2))/3)^2`

Explanation:

The centre of the circle can also be found as the intersection of two lines using linear equations as follows:

Let `A=(-3, -5)`, `B=(1, 3)` and `C=(5, -1)`

The midpoint of `AB` is `((-3+1)/2, (-5+3)/2) = (-1, -1)`

The slope of `AB` is `(3-(-5))/(1-(-3)) = 8/4 = 2`

So any perpendicular to `AB` has slope `-1/2`

The perpendicular through the midpoint of `AB` can be written:

`color(blue)(y+1 = -1/2 (x+1))`

The points on this line are equidistant from `A` and `B`.

The midpoint of `BC` is `((1+5)/2, (3-1)/2) = (3, 1)`

The slope of `BC` is `(-1-3)/(5-1) = (-4)/4 = -1`

So any perpendicular to `BC` has slope `-(1/-1) = 1`

The perpendicular through the midpoint of `BC` can be written:

`color(blue)(y-1 = x-3)`

The points on this line are equidistant from `B` and `C`.

Subtract the two equations of the perpendicular lines to find:

`2 = -1/2(x+1) - (x-3) =-3/2x+5/2`

Multiply both sides by `2` to get:

`4 = -3x+5`

Subtract `5` from both sides:

`-1 = -3x`

Divide both sides by `-3`:

`x = 1/3`

Then substitute this into the second equation to find `y = -5/3`

So the circle has equation `(x-1/3)^2+(y+5/3)^2 = r^2` for some radius `r`.

Substitute point `B` into this equation to find:

`r^2 = (1-1/3)^2 + (3+5/3)^2 =4/9+ 196/9 = 200/9`

So `r = sqrt(200/9) = (10 sqrt(2))/3`

So our equation in standard form is:

`(x-1/3)^2+(y+5/3)^2 = ((10 sqrt(2))/3)^2`