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

1 Answer
Dec 31, 2015

Answer:

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#