centre equals the midpoint = (5,-4)
equation:
(x-5)^2 + (y+4)^2 = r^2
plug in (3,-2)
4 + 4 = r^2 = 8
(x-5)^2 + (y+4)^2 = 8
find the equation of a circle. the endpoints of the diameter of a circle are (3,-2) and (7,-6).
4 answers
Those are endpoints of A diameter which is neither horizontal nor vertical. There are other possible diameter endpoint pairs for the same circle.
The midpoint of the given diameter is at (5,-4). You get that by averaging the two x's and the two y's, separately.
The length of the diameter is
D = sqrt(4^2 + 4^2) = sqrt32 = 4sqrt2
The radius is R = 2sqrt2
The equation of the circle is therefore
(x-5)^2 + (y+4)^2 = R^2 = 8
or
(x-5)^2/8 + (y+4)^2/8 = 1
Check: when x = 3 and y = -2
4/8 + 4/8 = 1
The midpoint of the given diameter is at (5,-4). You get that by averaging the two x's and the two y's, separately.
The length of the diameter is
D = sqrt(4^2 + 4^2) = sqrt32 = 4sqrt2
The radius is R = 2sqrt2
The equation of the circle is therefore
(x-5)^2 + (y+4)^2 = R^2 = 8
or
(x-5)^2/8 + (y+4)^2/8 = 1
Check: when x = 3 and y = -2
4/8 + 4/8 = 1
I got everything the same except the radius. How did you get 8?
The radius is not 8. It is the square root of 8.