Asked by Shredder
Find the equation of a circumcircle of a triangle whose testices are A (2, 3), B (5, 4) and C (3, 7)
Answers
Answered by
Steve
that's vertices, not testices.
The circumcenter lies at the intersection of the sides' perpendicular bisectors.
So, to start, find the equations of two of the bisectors.
AB: slope=1/3 midpoint is (7/2,7/2)
pb: (y-7/2) = (-3)(x-7/2)
AC: slope=4 midpoint is (5/2,5)
pb: (y-5) = (-1/4)(x-5/2)
the pb intersect at D:(67/22,107/22)
The distance AD is √2210/22
So, the circle is
(x-67/22)^2 + (y-107/22)^2 = 2210/484
see the graphs at
http://www.wolframalpha.com/input/?i=plot+y-3+%3D+%281%2F3%29%28x-2%29%2C+y-3+%3D+%284%29%28x-2%29%2C+y-4+%3D+%28-3%2F2%29%28x-5%29%2C+%28x-67%2F22%29^2+%2B+%28y-107%2F22%29^2+%3D+2210%2F484
The circumcenter lies at the intersection of the sides' perpendicular bisectors.
So, to start, find the equations of two of the bisectors.
AB: slope=1/3 midpoint is (7/2,7/2)
pb: (y-7/2) = (-3)(x-7/2)
AC: slope=4 midpoint is (5/2,5)
pb: (y-5) = (-1/4)(x-5/2)
the pb intersect at D:(67/22,107/22)
The distance AD is √2210/22
So, the circle is
(x-67/22)^2 + (y-107/22)^2 = 2210/484
see the graphs at
http://www.wolframalpha.com/input/?i=plot+y-3+%3D+%281%2F3%29%28x-2%29%2C+y-3+%3D+%284%29%28x-2%29%2C+y-4+%3D+%28-3%2F2%29%28x-5%29%2C+%28x-67%2F22%29^2+%2B+%28y-107%2F22%29^2+%3D+2210%2F484
Answered by
Akintayo
I need the working and answer
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