A triangle has corners at $(8 ,3 )$ , $(2 ,4 )$ , and $(7 ,2 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-04-18T13:08:10

The circle circumscribed on a triangle is the one that passes through the three vertices.

The general equation for a circle with center $(a,b)$ and squared radius $k$ is

$(x - a)^2 + (y-b)^2 = k$

Substituting our three points,

$(8 - a)^2 + (3 - b)^2 = k$

$(2 - a)^2 + (4- b)^2 = k$

$(7 -a )^2 + (2-b)^2 = k$

$a^2 + b^2 - 16 a - 6 b + 64 + 9 = k$

$a^2 + b^2 - 4a - 8b + 4 + 16 = k$

$a^2 + b^2 - 14 a - 4 b + 49 + 4 = k$

Subtracting pairs,

$-12 a + 2b + 53 = 0$

$-2 a -2 b + 20 = 0$

Adding,

$-14 a + 73 = 0$

$a = 73/14$

$12 a + 12 b - 120 = 0$

$14b - 67 = 0$

$b = 67/14$

$k = (2 - 73/14 )^2 + (4- 67/14)^2$

$k = 1/14^2 ( (28-73 )^2 + (56-67)^2 ) = 2146/14^2 = 1073/98$

Usually they pick nicer numbers for these problems. Our circle's area is $A = \pi r^2 = \pi k$

$A = frac { 1073 pi }{98}$

A triangle has corners at (8 ,3 )(8 ,3 ), (2 ,4 )(2 ,4 ), and (7 ,2 )(7 ,2 ). What is the area of the triangle's circumscribed circle?