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

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

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Answer 1 · 2017-01-30T02:59:35

The area of the circumscribed circle is $A = \frac{37 π}{2}$ $A = (37pi)/2$

Explanation:

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

$(2, 5) - (2, 5) \to (0, 0)$ $(2,5)-(2,5) to (0,0)$
$(9, 5) - (2, 5) \to (7, 0)$ $(9,5)-(2,5) to (7,0)$
$(3, 6) - (2, 5) \to (1, 1)$ $(3,6)-(2,5)to (1,1)$

Use the standard Cartesian form for the equation of a circle,

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x - h)^2 + (y - k)^2 = r^2$

and the 3 new points to write 3 equations:

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

Equation [1] reduces to, $h^{2} + k^{2} = r^{2} [4]$ $h^2 + k^2 = r^2" [4]"$

Substitute the left side of equation [4] into the right side of equations [2] and [3]:

${(7 - h)}^{2} + {(0 - k)}^{2} = h^{2} + k^{2} [5]$ $(7 - h)^2 + (0 - k)^2 = h^2 + k^2" [5]"$
${(1 - h)}^{2} + {(1 - k)}^{2} = h^{2} + k^{2} [6]$ $(1 - h)^2 + (1 - k)^2 = h^2 + k^2" [6]"$

Expand the squares on the left side of equations [5] and [6], using the pattern ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a - b)^2 = a^2 -2ab + b^2$ :

$49 - 14 h + h^{2} + k^{2} = h^{2} + k^{2} [7]$ $49 - 14h+ h^2 + k^2 = h^2 + k^2" [7]"$
$1 - 2 h + h^{2} + 1 - 2 k + k^{2} = h^{2} + k^{2} [8]$ $1 - 2h+h^2 + 1 - 2k+ k^2 = h^2 + k^2" [8]"$

The $h^{2} and k^{2}$ $h^2 and k^2$ terms cancel in both equations:

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

Equation [9] allows us to solve for h:

$h = \frac{7}{2}$ $h = 7/2$

Substitute $\frac{7}{2}$ $7/2$ for h into equation [10] and solve for k:

$1 - 2 (\frac{7}{2}) + 1 - 2 k = [10]$ $1 - 2(7/2) + 1 - 2k =" [10]"$

$k = \frac{5}{2}$ $k = 5/2$

Use equation [4] to solve for $r^{2}$ $r^2$ :

$r^{2} = {(\frac{7}{2})}^{2} + {(\frac{5}{2})}^{2}$ $r^2 = (7/2)^2 + (5/2)^2$

$r^{2} = \frac{74}{4} = \frac{37}{2}$ $r^2 = 74/4 = 37/2$

The area of a circle is $A = π r^{2}$ $A = pir^2$

The area of the circumscribed circle is $A = \frac{37 π}{2}$ $A = (37pi)/2$

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