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)?

1 Answer
Dec 14, 2015

Answer:

#(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]}