Question

How do you find the equation of a circle with diameter has endpoints (-2, 3) and (4, -1)?

Answer 1

`(x-1)^2+(y-1)^2 = 13`

Explanation:

The radius of the circle is half of the length of the diameter, so half the distance between these two endpoints:

`1/2 sqrt((4-(-2))^2+((-1)-3)^2) = 1/2 sqrt(36+16) = 1/2 sqrt(52) = sqrt(13)`

The centre of the circle is the midpoint of the diameter:

`(((-2)+4)/2, (3+(-1))/2) = (1, 1)`

The equation of a circle with centre `(h, k)` and radius `r` may be written:

`(x-h)^2+(y-k)^2 = r^2`

So in our case:

`(x-1)^2+(y-1)^2 = 13`

graph{((x-1)^2+(y-1)^2-13)((x-1)^2+(y-1)^2-0.02)((x+2)^2+(y-3)^2-0.02)((x-4)^2+(y+1)^2-0.02) = 0 [-10, 10, -5, 5]}