The circumcenter of a triangle is the center of a circumference where the three vertex are included. So basically we must find the circumference that passes through points O, V and W. The equation of a circumference of a radius r and a central point (a,b) is:
[tex](x-a)^2+(y-b)^2=r^2[/tex]We have three points which give us three pairs of (x,y) values that we can use to build three equations for a, b and r. Using point O=(6,5) we get:
[tex](6-a)^2+(5-b)^2=r^2[/tex]Using V=(0,13) we get:
[tex](0-a)^2+(13-b)^2=r^2[/tex]And using W=(-3,0) we get:
[tex](-3-a)^2+(0-b)^2=r^2[/tex]So we have a system of three equations and we must find three variables: a, b and r. All equations have r^2 at their right side. This means that we can take the left sides and equalize them. Let's do this with the second and third equation:
[tex]\begin{gathered} (0-a)^2+(13-b)^2=(-3-a)^2+(0-b)^2 \\ a^2+(13-b)^2=(-3-a)^2+b^2 \end{gathered}[/tex]If we develop the squared terms:
[tex]a^2+b^2-26b+169=a^2+6a+9+b^2[/tex]Then we substract a^2 and b^2 from both sides:
[tex]\begin{gathered} a^2+b^2-26b+169-a^2-b^2=a^2+6a+9+b^2-a^2-b^2 \\ -26b+169=6a+9 \end{gathered}[/tex]We substract 9 from both sides:
[tex]\begin{gathered} -26b+169-9=6a+9-9 \\ -26b+160=6a \end{gathered}[/tex]And we divide by 6:
[tex]\begin{gathered} \frac{-26b+160}{6}=\frac{6a}{6} \\ a=-\frac{13}{3}b+\frac{80}{3} \end{gathered}[/tex]Now we can replace a with this expression in the first equation:
[tex]\begin{gathered} (6-a)^2+(5-b)^2=r^2 \\ (6-(-\frac{13}{3}b+\frac{80}{3}))^2+(5-b)^2=r^2 \\ (\frac{13}{3}b-\frac{62}{3})^2+(5-b)^2=r^2 \end{gathered}[/tex]We develop the squares:
[tex]\begin{gathered} (\frac{13}{3}b-\frac{62}{3})^2+(5-b)^2=r^2 \\ \frac{169}{9}b^2-\frac{1612}{9}b+\frac{3844}{9}+b^2-10b+25=r^2 \\ \frac{178}{9}b^2-\frac{1702}{9}b+\frac{4069}{9}=r^2 \end{gathered}[/tex]So this expression is equal to r^2. This means that is equal
