Asked by Anonymous
Find the center of the circle that can be circumscribed about EFG with E(4, 4), F(4, 2), and G(8, 2
Answers
Answered by
Steve
If the center is at (h,k) and the radius is r, then you can solve
(4-h)^2+(4-k)^2 = r^2
(4-h)^2+(2-k)^2 = r^2
(8-h)^2+(2-k)^2 = r^2
or, subtracting judiciously,
(4-k)^2 = (2-k)^2
(8-h)^2 = (2-h)^2
That gives (h,k) = (5,3)
(4-h)^2+(4-k)^2 = r^2
(4-h)^2+(2-k)^2 = r^2
(8-h)^2+(2-k)^2 = r^2
or, subtracting judiciously,
(4-k)^2 = (2-k)^2
(8-h)^2 = (2-h)^2
That gives (h,k) = (5,3)
Answered by
Dude
The possible answer choices:
(3, 6)
(6, 3)
(4, 2)
(4, 4)
(3, 6)
(6, 3)
(4, 2)
(4, 4)
Answered by
Dude
I found the answer to be (6,3) using planetcalc.com/8116/