Question

How do you find general form of circle with endpoints of a diameter at the points (4, 3) and (0, 1)?

Answer 1

General Form: `(x-2)^2+(y-2)^2=5`
or `x^2-4x+y^2-4y+3=0`

Explanation:

The center of the circle is the middle point of the diameter
`x_c = (4+0)/2=2` and `y_c=(3+1)/2`. Then C(2,2).

The distance between the center of the circle and to a endpoint of a diameter is equal to the radius:
`R=sqrt((4-2)^2+(3-2)^2)=sqrt(4+1)=sqrt(5)`

The formula of the general form of the circle is
`(x-x_c)^2+(y-y_c)^2=R^2`

Then, in case, the equation of the circle is
`(x-2)^2+(y-2)^2=5`