Question

If the endpoints of the diameter of a circle are (6,3) and (2,1), what is the standard form of the circle?

Answer 1

`r= √(5)`
`(8)/(2) , (4)/(2)`
`distance= √((x-x)^2+(y-y)^2) = √((2-6)^2+(1-3)^2) = 2√(5)`
`r= √(5)`
through midpoint we find and b to be (4,2).

`(x-4)^2+(y-2)^2=5`