Let the center of the circum-circle be (h,k)(h,k) and its radius be rr. Then, the equation of the circle is
(x-h)^2+(y-k)^2 = r^2(x−h)2+(y−k)2=r2
Since this circle must pass through the three vertices we have
(5-h)^2+(8-k)^2 = (2-h)^2+(7-k)^2 = (7-h)^2+(3-k)^2 = r^2(5−h)2+(8−k)2=(2−h)2+(7−k)2=(7−h)2+(3−k)2=r2
From (5-h)^2+(8-k)^2 = (2-h)^2+(7-k)^2(5−h)2+(8−k)2=(2−h)2+(7−k)2 we get
25-10h+h^2 +64-16k+k^2 = 4-4h+k^2+49-14k+k^225−10h+h2+64−16k+k2=4−4h+k2+49−14k+k2
so that
color(red)(6h + 2k = 36)6h+2k=36
Similarly, (2-h)^2+(7-k)^2 = (7-h)^2+(3-k)^2(2−h)2+(7−k)2=(7−h)2+(3−k)2 gives us
4-4h+h^2 +49-14k+k^2 = 49-14h+h^2+9-6k+k^2 4−4h+h2+49−14k+k2=49−14h+h2+9−6k+k2 which simplifies to
color(red)(10h-8k = 5)10h−8k=5
Solving the two equations simultaneously gives
h = 149/34,quad k = 165/34
This leads to
r^2 ~~ 1.29
so that the area of the circumscribed circle is
pi r^2 ~~ 32.3
