find the coordinates of the circumcenter of a triangle ABC whose vertices are (4,6) (0,4) (6,2) resp

Let P(x,y) be the circumcentre of the triangle ABC whose vertices are A(4,6), B(0,4) and C(6,2).
Then,
$$\begin{gathered} CP=AP=BP \\ Now, \\ CP=AP \\ \Rightarrow C{P}^2=A{P}^2 \\ \Rightarrow {\left(x-6\right)}^2+{\left(y-2\right)}^2={\left(x-4\right)}^2+{\left(y-6\right)}^2 \\ \Rightarrow \left(x^2+36-12x\right)+\left(y^2+4-4y\right)=\left(x^2+16-8x\right)+\left(y^2+36-12y\right) \\ \Rightarrow 40-12x-4y=-8x-12y+52 \\ \Rightarrow -4x+8y-12=0 \\ \Rightarrow x-2y+3=0 .....\left(1\right) \end{gathered}$$
$$\begin{gathered} CP=BP \\ \Rightarrow C{P}^2=B{P}^2 \\ \Rightarrow {\left(x-6\right)}^2+{\left(y-2\right)}^2={\left(x-0\right)}^2+{\left(y-4\right)}^2 \\ \Rightarrow \left(x^2+36-12x\right)+\left(y^2+4-4y\right)=\left(x^2\right)+\left(y^2+16-8y\right) \\ \Rightarrow 40-12x-4y=16-8y \\ \Rightarrow -12x+4y+24=0 \\ \Rightarrow 3x-y-6=0 .....\left(2\right) \end{gathered}$$
Multiply by 2 in equation (2) and then subtracting equation (1) and (2), we get,
$$\begin{gathered} -5x+15=0 \\ \Rightarrow x=3 \end{gathered}$$
Substitute value of x in equation (1),
$$\begin{gathered} 3-2y+3=0 \\ \Rightarrow y=3 \\ Thus, the coordinates of the circumcentre are \left(3,3\right). \end{gathered}$$