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\times4+4y=9$$4y=9-8$$y=\frac{1}{4}=0.25$

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

$(4,0.25)$