Question

For the following, can you help me find the circumcenter within the diagram

Answer 1

First draw the diagram.

For circumcenter , PB=PA, PA=PC

Apply PA=PB

`(2-x)^2+(0-y)^2=(3-x)^2+(2-y)^2`
`4+x^2-4x+y^2=9+x^2-6x+4+y^2-4y`
`-4x=9-6x-4y`
`2x+4y=9`

For PA=PC.

`(2-x)^2+(0-y)^2=(6-x)^2+(0-y)^2`
`4+x^2-4x+y^2=36+x^2-12x+y^2`
`4-4x=36-12x`
`8x=32`
`x=4`

Substitute the value of x in the equation above to find the y.

`2*4+4y=9`
`4y=9-8`
`y=(1)/(4)=0.25`

Hence the circumcenter of the triangle which is the coordinate P is

`(4,0.25)`