Analítica, fundamentos da matemática

Determinar a circunferencia circunscrita ao triângulo de vértices A(5,4), B (6,1), C (-3,-2)

R = (X-1)^2 + (Y-1)^2 = 25

Tentei achar o centro fazendo ponto médio de AB (= ponto D), e depois ponto médio de CD mas, sem sucesso... se puderem me ajudar fico grato

[latex]\\ A = (5,4)\;\;\;\;\;\;\;\;\;\;B=(6,1)\\\\ C=(-3,-2)\;\;\; D = (m,n)\\\\\\ d_{AD} = d_{BD} = d_{CD} = R\\\\\\ d_{AD}^2=(5-m)^2+(4-n)^2=R^2\\\\ d_{BD}^2=(6-m)^2+(1-n)^2=R^2\\\\ d_{CD}^2=(-3-m)^2+(-2-n)^2=R^2\\\\\\\\ \left\{\begin{matrix} m^2 - 10 m + n^2 - 8 n + 41=R^2\mbox{ (I)}\\\\ m^2 - 12 m + n^2 - 2 n + 37=R^2\mbox{ (II)}\\\\ m^2 + 6 m + n^2 + 4 n + 13=R^2\;\;\mbox{ (III)} \end{matrix}\right.\\\\\\\\ \left\{\begin{matrix} m^2 - 10 m + n^2 - 8 n + 41=R^2\mbox{ (I)}\\\\ m - 3 n + 2=0\mbox{ (I - II)}\\\\ 4 m +3 n -7=0\;\;\mbox{ (I - III)} \end{matrix}\right. \\\\\\\\ \left\{\begin{matrix} R^2 = 25\\\\ m = 1\\\\ n = 1 \end{matrix}\right.\;\;\Rightarrow\;\;\boxed{(x-1)^2+(y-1)^2 = 25}[/latex]