A triangle has corners at $(4 ,7 )$ , $(1 ,3 )$ , and $(6 ,5 )$ . What is the area of the triangle's circumscribed circle?

Question

1299 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-21T10:17:34

$frac{725 pi}{98}$

The circumcised circle is just the circle that has those three points on it. We need the squared radius, so let's solve for the circle's equation.

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

Each point gives us an equation:

$(4-a)^2+(7-b)^2=r^2$
$(1-a)^2+(3-b)^2=r^2$
$(6-a)^2+(5-b)^2=r^2$

Expanding,

$16 - 8a + a^2 + 49 - 14b + b^2 = r^2$
$1 - 2a + a^2 + 9 - 6b + b^2 = r^2$
$36 - 12a + a^2 + 25 - 10b + b^2 =r ^2$

Subtracting pairs,

$65 - 10 - 8 a + 2a - 14b + 6b = 0$
$55= 6a + 8 b$
$65 - 61- 8 a + 12 a - 14 b + 10 b = 0$
$4 = -4 a + 4 b$
$1 = -a + b$
$8 = -8 a + 8 b$

Subtracting,

$55 - 8 = 6a - -8a$
$a = 47/14$
$b=a+1=61/14$

What we're really after is $r^2$ . We substitute any one of the points:

$r^2 = (6 - 47/14)^2 + (5-61/14)^2$
$= frac{1}{14^2} ( ( 6(14)-47)^2 + (5(14) - 61)^2 )$

$r^2 = 725/98$

The area we seek is

$A = \pi r^2 = frac{725 pi}{98}$

A triangle has corners at (4 ,7 )(4 ,7 ), (1 ,3 )(1 ,3 ), and (6 ,5 )(6 ,5 ). What is the area of the triangle's circumscribed circle?