Need some help. how do you find the equation of a circle with endpoints of the diameter at (-1,1) and (7,9)?

Question

1388 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-11T03:00:50

${(x - 3)}^{2} + {(y - 5)}^{2} = 32$ $(x-3)^2 + (y-5)^2 = 32$

Urgent!

The center is the midpoint of the diameter:

$(a, b) = (\frac{- 1 + 7}{2}, \frac{1 + 9}{2}) = (3, 5)$ $(a,b) = ({-1 + 7}/2, {1+9}/2 ) = (3,5)$

The squared radius is the squared distance from the center to either endpoint:

$r^{2} = {(3 - - 1)}^{2} + {(5 - 1)}^{2} = 32$ $r^2 = (3 - -1)^2 + (5-1)^2 = 32$

The general equation for a circle with center $(a, b)$ $(a,b)$ and radius $r$ $r$ is

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2 + (y-b)^2 = r^2$

${(x - 3)}^{2} + {(y - 5)}^{2} = 32$ $(x-3)^2 + (y-5)^2 = 32$