What's The Radius?

Calculate the sides of the triangle using distance formula. and use the relation Area of triangle = (abc/4R) Area of triangle = sq root (s (s-a)(s-b)(s-c)) where s=(a+b+c)/2

a,b,c sides of triangle R is circumradius of triangle

Let the coordinates of the center of the circumcircle be $O(x,y)$ and its radius $R$ .

Then, we have:

$\begin{cases} R^2 & = & (x-400)^2+(y-155)^2 & ...(1) \\ R^2 & = & (x-650)^2+(y-550)^2 & ...(2) \\ R^2 & = & (x-370)^2+(y-636)^2 & ...(3) \end{cases}$

$\begin{cases} R^2 & = & x^2-800x+160000+y^2-310y+24025 & ...(1) \\ R^2 & = & x^2-1300x+422500+y^2-1100y+302500 & ...(2) \\ R^2 & = & x^2-740y+136900+y^2-1272y+404496 & ...(3) \end{cases}$

$Eq. 1 - Eq. 2$ and $Eq. 3 - Eq. 2$ :

$\begin{cases} 0 & = & 500x-262500+790y-278475 & ...(4) \\ 0 & = & 560x-285600-172y+101996 & ...(5) \end{cases}$

$\begin{cases} 500x+790y & = & 540975 & ...(4) \\ 560x-172y & = & 183604 & ...(5) \end{cases}$

$Eq. 4/790$ and $Eq. 5/175$ :

$\begin{cases} 0.632911392x+y & = & 684.778481 & ...(6) \\ 3.255813953x-y & = & 1067.465116 & ...(7) \end{cases}$

$Eq. 6 + Eq. 7$ : $\quad 3.888725346x = 1752.243597\quad \Rightarrow x = 450.5958743$

Substituting value of $x$ in $Eq. 6$ : $\quad \Rightarrow y = 399.5912188$

Substituting values of $x$ and $y$ in $Eq.1$ :

$R^2 = (450.5958743-400)^2+(399.5912188-155)^2$

$\quad \space = (50.5958743-400)^2+(244.5912188-155)^2$

$\quad \space = 2559.9425+59824.8643$

$\quad \space =62384.8068$

$\Rightarrow R = \sqrt {62384.8068} = \boxed {249.77}$

We can use the extended sine rule. a/sin(A) = b/sin(B) =c/sin(C) = 2R where R is the circumradius. Since we are given the coordinates, we can use Pythagoras to find out the length of the sides. We can then use the cosine rule to find out one of the angles.

$Use\ midpoint\ and\ slope\ formulii\ for\ AB\ to\ get\ F\ and\ m_c.\\ Similarly\ get\ E\ and\ m_b\ for \ AC.\\ Line \ from\ D\ and\ slope\ \dfrac {-1}{m_c},\ intersect\ line \ from\ F\ and\ slope\ \dfrac {-1}{m_b}\ \ at\ O.\\ Use \ length \ formula \ to \ find \ the \ length \ of\ AO=required \ R=249.77.$

Kudos to Sujay Motani for his really elegant solution, unlike my rather tedious solution.

Let $\theta = \angle BAC$ and the radius be $r$ .

By the cosine rule, $\cos \theta = \frac{AB^{2}+AC^{2}-BC^{2}}{2 \times AB \times AC}$ .

Similarly, $\cos 2\theta = \frac{r^{2}+r^{2}-BC^{2}}{2 \times r \times r} = \frac{2r^{2}-BC^{2}}{2r^{2}}$ .

Combining the two equations gives

$2\cos^{-1} (\frac{AB^{2}+AC^{2}-BC^{2}}{2 \times AB \times AC}) = \cos^{-1} (\frac{2r^{2}-BC^{2}}{2r^{2}})$ .

After a bit of manipulation, we get

$r = \frac{BC}{\sqrt{2-2\cos (2\cos^{-1} (\frac{AB^{2}+AC^{2}-BC^{2}}{2 \times AB \times AC}))}}$ .

Plugging in the values of $AB, AC$ and $BC$ by using the distance formula $d = \sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$ , we get $r \approx \boxed{249.77}$ .