A triangle has corners at $(5, 7)$ $(5 ,7 )$ , $(2, 1)$ $(2 ,1 )$ , and $(1, 6)$ $(1 ,6 )$ . What is the area of the triangle's circumscribed circle?

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A triangle has corners at $(5, 7)$ $(5 ,7 )$ , $(2, 1)$ $(2 ,1 )$ , and $(1, 6)$ $(1 ,6 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-02-02T14:25:41

Area of circumcircle $A_{c} = 35.5878$ $A_c = 35.5878$

Explanation:

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$M_{3} = \frac{5 + 2}{2}, \frac{7 + 1}{2} = (\frac{7}{2}, 4)$ $M_3 = (5+2)/2, (7+1)/2 = (7/2, 4)$

Slope of A1A2 = (7-1)/(5-2) = 2#

Slope of perpendicular bisector thro $M_{3} = - \frac{1}{2}$ $M_3= -1/2$

Equation of perpendicular bisector through $M_{3}$ $M_3$ is

$y - 4 = (- \frac{1}{2}) (x - (\frac{7}{2}))$ $y - 4 = (-1/2) (x - (7/2))$

$4 y - 16 = - 2 x + 7$ $4y - 16 = -2x + 7$

$4 y + 2 x = 23$ $4y + 2x = 23$ Eqn (1)

$M_{1} = \frac{2 + 1}{2}, \frac{1 + 6}{2} = (\frac{3}{2}, \frac{7}{2})$ $M_1 = (2+1)/2, (1+6)/2 = (3/2, 7/2)$

Slope of A2A3 = (6-1) / (1-2) = -6#

Slope of perpendicular bisector through $M_{1} = - \frac{1}{- 6} = \frac{1}{6}$ $M_1 = -1/-6 = 1/6$

Equation of perpendicular bisector through $M_{1}$ $M_1$ is

$y - (\frac{7}{2}) = (\frac{1}{6}) (x - (\frac{3}{2}))$ $y - (7/2) = (1/6) (x - (3/2))$

$12 y - 42 = 2 x - 3$ $12y - 42 = 2x - 3$

$12 y - 2 x = 39$ $12y - 2x = 39$ Eqn (2)

Solving equations (1), (2) we get coordinates of circumcenter O.

$O (\frac{15}{4}, \frac{31}{8})$ $O (15/4, 31/8)$

Circumcircle radius $R_{o} = \sqrt{{((\frac{15}{4}) - 2)}^{2} + {((\frac{31}{8}) - 1)}^{2}} = 3.3657$ $R_o = sqrt(((15/4)-2)^2 + ((31/8) - 1)^2) = 3.3657$

Area of circumcircle $A_{c} = π R_{o}^{2} = π \cdot {(3.3657)}^{2} = 35.5878$ $A_c = pi R_o^2 = pi * (3.3657)^2 = 35.5878$

Answer 2 · 2018-05-08T15:21:42

Circumcircle area $= \frac{1105 π}{98}$ $= {1105 pi}/98$

Explanation:

I worked out the general case here .

I'm going to ignore the details of the general answer, and state the main result as: The squared radius of the circumcircle equals the product of the squared sides of the triangle divided by sixteen times the squared area of the triangle. Given triangle with sides $a, b, c$ $a,b,c$ and area $A$ $A$ , the circumradius $r$ $r$ satisfies

$r^{2} = \frac{a^{2} b^{2} c^{2}}{16 A^{2}}$ $r^2 = {a^2b^2c^2}/{16A^2}$

It's tempting to take the square root, but experience shows much smoother sailing if we don't. We can get the area from the coordinates using the Shoelace Theorem, or get $16 A^{2}$ $16A^2$ directly from the squared sides using Archimedes' Theorem:

$16 A^{2} = 4 a^{2} b^{2} - {(c^{2} - a^{2} - b^{2})}^{2}$ $16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2$

Since we need the squared sides for the numerator anyway, let's do it this way. We'll label the vertices $A (5, 7),$ $A(5,7),$ $B (2, 1),$ $B(2,1),$ $C (1, 6)$ $C(1,6)$ and calculate

$a^{2} = B C^{2} = {(2 - 1)}^{2} + {(1 - 6)}^{2} = 1^{2} + 5^{2} = 26$ $a^2 = BC^2 = (2-1)^2+(1-6)^2=1^2+5^2=26$

$b^{2} = A C^{2} = 4^{2} + 1^{2} = 17$ $b^2 = AC^2 = 4^2+1^2=17$

$c^{2} = A B^{2} = 3^{2} + 6^{2} = 45$ $c^2 = AB^2 = 3^2+6^2=45$

$16 A^{2} = 4 (26) (17) - {(45 - 26 - 17)}^{2} = 1764$ $16 A^2 = 4(26)(17)-(45-26-17)^2 = 1764$

$r^{2} = \frac{a^{2} b^{2} c^{2}}{16 A^{2}} = \frac{(26) (17) (45)}{1764} = \frac{1105}{98}$ $r^2 = {a^2b^2c^2}/{16A^2} = { (26)(17)(45) }/ 1764 = 1105/98$

Circumcircle area $= π r^{2} = \frac{1105 π}{98}$ $= pi r^2 = {1105 pi}/98$

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A triangle has corners at $(5, 7)$ $(5 ,7 )$ , $(2, 1)$ $(2 ,1 )$ , and $(1, 6)$ $(1 ,6 )$ . What is the area of the triangle's circumscribed circle?

2 Answers

Explanation:

Explanation:

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A triangle has corners at (5,7)(5 ,7 ), (2,1)(2 ,1 ), and (1,6)(1 ,6 ). What is the area of the triangle's circumscribed circle?

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Explanation:

Explanation:

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A triangle has corners at $(5, 7)$ $(5 ,7 )$ , $(2, 1)$ $(2 ,1 )$ , and $(1, 6)$ $(1 ,6 )$ . What is the area of the triangle's circumscribed circle?