Question

How do you find the general form of the equation of this circle with end points of a diameter at (1,4) and (-3,2)?

Answer 1

`(x+1)^2+(y-3)=5`

Explanation:

If the end points of a diameter of a circle are at `(1,4)` and `(-3,2)`
then the center of the circle is at
`color(white)("XXX")((1+(-3))/2,(4+2)/2)=(-1,3)`
and the radius of the circle is
`color(white)("XXX")sqrt((1-(-1))^2+(4-3)^2) = sqrt(5)`

The general equation of a circle with center `(a,b)` and radius `r` is
`color(white)("XXX")(x-a)^2+(y-b)^2=r^2`

The required circle has an equation of
`color(white)("XXX")(x+1)^2+(y-3)^2=5`
graph{(x+1)^2+(y-3)^2=5 [-5.47, 4.396, 0.33, 5.264]}