A triangle has corners at $(2 ,4 )$ , $(3 ,6 )$ , and $(4 ,7 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-04-21T13:11:41

$65/2 pi$

Just did one just like this.

The circumscribed circle is just the circle with the three vertices on it. The general equation is

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

Substituting three points,

$(2-a)^2 + (4-b)^2=r^2$
$(3-a)^2+(6-b)^2=r^2$
$(4-a)^2+(7-b)^2=r^2$

Expanding,

$20 = 4a + 8b + r^2-a^2-b^2$
$45 = 6a + 12 b + r^2-a^2-b^2$
$65 = 8a + 14b + r^2-a^2-b^2$

Subtracting pairs,

$25 = 2a + 4b$
$20 = 2a + 2b$

Subtracting,

$5 = 2b$
$b=5/2$
$a = 10 - b = 15/2$

Substituting,

$r^2 = (2-15/2)^2 + (4-5/2)^2= 65 /2$

So the circle's area is

$A = \pi r^2 = 65/2 pi$

A triangle has corners at (2 ,4 )(2 ,4 ), (3 ,6 )(3 ,6 ), and (4 ,7 )(4 ,7 ). What is the area of the triangle's circumscribed circle?