well the middle is halfway between
(8/2 , -12/2) = (4,-6) center
so
(x-4)^2 + (y+6)^2 = r^2
try a point like (1,-7)
(-3)^2 + (-7+6)^2 = r^2
9 + 1 = r^2 so r = sqrt 10 =
(x-4)^2 +(y+6)^2 = 10
========================check with other point (7,-5)
(3)^2 + (1)^2 = 9 + 1 = 10 good, checks
Find the standard form of the equation of the circle with the given characteristics.
Endpoints of a diameter: (1, −7) and (7, −5)
1 answer
1 answer