A triangle has corners at $(5 ,8 )$ , $(2 ,7 )$ , and $(7 ,3 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-03-05T17:33:18

$~~ 32.3$

Let the center of the circum-circle be $(h,k)$ and its radius be $r$ . Then, the equation of the circle is

$(x-h)^2+(y-k)^2 = r^2$

Since this circle must pass through the three vertices we have

$(5-h)^2+(8-k)^2 = (2-h)^2+(7-k)^2 = (7-h)^2+(3-k)^2 = r^2$

From $(5-h)^2+(8-k)^2 = (2-h)^2+(7-k)^2$ we get

$25-10h+h^2 +64-16k+k^2 = 4-4h+k^2+49-14k+k^2$

so that

$color(red)(6h + 2k = 36)$

Similarly, $(2-h)^2+(7-k)^2 = (7-h)^2+(3-k)^2$ gives us

$4-4h+h^2 +49-14k+k^2 = 49-14h+h^2+9-6k+k^2$ which simplifies to

$color(red)(10h-8k = 5)$

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
$pi r^2 ~~ 32.3$

A triangle has corners at (5 ,8 )(5 ,8 ), (2 ,7 )(2 ,7 ), and (7 ,3 )(7 ,3 ). What is the area of the triangle's circumscribed circle?