Question

Show that the centre of any circle which passes through (-1, 1) and (1. 5) lies on the line x+2y-6=0 ?

Answer 1

The Cartesian equation for a circle is:

`(x-h)^2+(y-k)^2=r^2" [1]"`

where `(x,y)` is any point on the circle, `(h,k)` is the center-point and `r` is the radius.

Substitute the points `(-1,1)` and `(1,5)` into equation [1]:

`(-1-h)^2+(1-k)^2=r^2" [2]"`
`(1-h)^2+(5-k)^2=r^2" [3]"`

Expand the squares of both equations:

`1+2h+h^2+1-2k+k^2=r^2" [4]"`
`1-2h+h^2+25-10k+k^2=r^2" [5]"`

Subtract equation [4] from equation [5]:

`1-2h+h^2+25-10k+k^2=r^2" [5]"`
`ul(1+2h+h^2+1-2k+k^2=r^2)" [4]"`
`0-4h+0h^2+24-8k+0k^2=0r^2`

Combine like terms:

`-4h-8k+24=0`

Divide both sides of the equation by -4:

`h+2k-6=0`

This is the line `x+2y-6=0` evaluated at the point `(h,k)`

Q.E.D.