Question

Find the equation of the circle passing through (2,3) (6,1) & (4,-3)?

Answer 1

`color(blue)((x-3)^2+y^2=10`

Explanation:

The standard equation of a circle is given as:

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

Where:

`bbh` and `bbk` are the `bbx` and `bby` co-ordinates of the centre respectively and `bbr` is the radius.

We are given 3 points on the circumference of the given circle. This enables us to form 3 independent equations. We need 3 equations, since we are solving for 3 variables, `bbh,bbk` and `bbr`

Using point `(2,3)`:

`(2-h)^2+(3-k)^2=r^2 \ \ \ [1]`

Using point `(6,1)`:

`(6-h)^2+(1-k)^2=r^2 \ \ \ [2]`

Using point `(4,-3)`

`(4-h)^2+(-3-k)^2=r^2 \ \ \ [3]`

Solving simultaneously:

Subtract `[2]` from `[1]`:

`4-36+8h+8-4k=0`

`8h-4k-24=0 \ \ \ \ [4]`

Subtract `[3]` from `[2]`

`20-4h-8-8k=0`

`12-4h-8k=0 \ \ \ [5]`

From `[5]`

`h=3-2k`

Substituting in `[4]`

`8(3-2k)-4k-24=0=>k=0`

Substituting in `[5]`

`12-4h-8(0)=0=>h=3`

Centre is `(h,k)=(3,0)`

Using this in `[1]` with point `(2,3)`

You can use `[1] ,[2] or [3]` here and any of the 3 points.

`(2-3)^2+(3-0)^2=r^2`

`1+9=r^2=>r^2=10`

So the given equation is:

`color(blue)((x-3)^2+y^2=10`

PLOT: