A triangle has corners at $(9, 5)$ $(9 ,5 )$ , $(2, 1)$ $(2 ,1 )$ , and $(3, 6)$ $(3 ,6 )$ . What is the area of the triangle's circumscribed circle?

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A triangle has corners at $(9, 5)$ $(9 ,5 )$ , $(2, 1)$ $(2 ,1 )$ , and $(3, 6)$ $(3 ,6 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2017-01-06T20:30:21

Shift ALL the points so that one is the origin. Use the standard Cartesian form for the equation of a circle and the new points to write 3 equations. Use the 3 equations to solve for $r^{2}$ $r^2$ .

Explanation:

Shift all 3 points so that one of them is the origin:

$(2, 1) - (2, 1) = (0, 0)$ $(2,1) - (2,1) = (0,0)$
$(9, 5) - (2, 1) = (7, 4)$ $(9,5) - (2,1) = (7,4)$
$(3, 6) - (2, 1) = (1, 5)$ $(3,6) - (2,1) = (1,5)$

This is the standard Cartesian form for the equation of a circle:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2} [1]$ $(x - h)^2 + (y - k)^2 = r^2" [1]"$

Use the new points to write 3 equations:

${(0 - h)}^{2} + {(0 - k)}^{2} = r^{2} [2]$ $(0 - h)^2 + (0 - k)^2 = r^2" [2]"$
${(7 - h)}^{2} + {(4 - k)}^{2} = r^{2} [3]$ $(7 - h)^2 + (4 - k)^2 = r^2" [3]"$
${(1 - h)}^{2} + {(5 - k)}^{2} = r^{2} [4]$ $(1 - h)^2 + (5 - k)^2 = r^2" [4]"$

Expand the squares:

$h^{2} + k^{2} = r^{2} [5]$ $h^2 + k^2 = r^2" [5]"$
$49 - 14 h + h^{2} + 16 - 8 k + k^{2} = r^{2} [6]$ $49 - 14h + h^2 + 16 - 8k + k^2 = r^2" [6]"$
$1 - 2 h + h^{2} + 25 - 10 k + k^{2} = r^{2} [7]$ $1 - 2h + h^2 + 25 - 10k + k^2 = r^2" [7]"$

Subtract equation [6] from equation [5] and equation [7] from equation [5]:

$- 49 + 14 h - 16 + 8 k = 0 [8]$ $-49 + 14h - 16 + 8k = 0" [8]"$
$- 1 + 2 h - 25 + 10 k = 0 [9]$ $-1 + 2h - 25 + 10k = 0" [9]"$

Collect the constant terms into a single term on the right:

$14 h + 8 k = 65 [10]$ $14h + 8k = 65" [10]"$
$2 h + 10 k = 26 [11]$ $2h + 10k = 26" [11]"$

Multiply equation [11] by -7 and add to equation [10]:

$- 62 k = - 117$ $-62k = -117$

$k = \frac{117}{62}$ $k = 117/62$

Substitute the value for k into equation [11] and solve for h:

$2 h + \frac{1170}{62} = 26 [11]$ $2h + 1170/62 = 26" [11]"$

$h = \frac{221}{62}$ $h = 221/62$

Use equation [5] to solve for $r^{2}$ $r^2$ :

$r^{2} = h^{2} + k^{2} [5]$ $r^2 = h^2 + k^2" [5]"$

$r^{2} = {(\frac{221}{62})}^{2} + {(\frac{117}{62})}^{2}$ $r^2 = (221/62)^2 + (117/62)^2$

$r^{2} = \frac{62530}{3844} = \frac{31265}{1922}$ $r^2 = 62530/3844 = 31265/1922$

The area of the circle is:

$A = \frac{31265 π}{1922}$ $A = (31265pi)/1922$

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A triangle has corners at $(9, 5)$ $(9 ,5 )$ , $(2, 1)$ $(2 ,1 )$ , and $(3, 6)$ $(3 ,6 )$ . What is the area of the triangle's circumscribed circle?

1 Answer

Explanation:

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A triangle has corners at (9,5)(9 ,5 ), (2,1)(2 ,1 ), and (3,6)(3 ,6 ). What is the area of the triangle's circumscribed circle?

1 Answer

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A triangle has corners at $(9, 5)$ $(9 ,5 )$ , $(2, 1)$ $(2 ,1 )$ , and $(3, 6)$ $(3 ,6 )$ . What is the area of the triangle's circumscribed circle?