How do you write the equation of the circle where C(1,-3) and D(-3,7) are the endpoints of a diameter?

1 Answer
May 9, 2016

Answer:

#(x+1)^2+(y-2)^2=29#

Explanation:

The midpoint of the diameter #CD# must be the center of the circle:
#color(white)("XXX") #center is at #(c_x,c_y)=((1+(-3))/2,((-3)+7)/2)=(-1,2)#

The length of the diameter is given by the Pythagorean Theorem:
#color(white)("XXX") abs(CD)=sqrt((1-(-3))^2+(-3-7)^2)#

#color(white)("XXX") =sqrt(116)#

#color(white)("XXX") =2sqrt(29)#

which implies that the radius of the circle is
#color(white)("XXX") r=sqrt(29)#

The general formula for a circle with center #(c_x,c_y)# and radius #r# is
#color(white)("XXX") (x-c_x)^2+(y-c_y)^2=r^2#

In this case:
#color(white)("XXX") (x+1)^2+(y-2)^2=29#

For verification purposes here is the graph of this circle equation with the given diameter endpoints:
graph{((x+1)^2+(y-2)^2-29)((x-1)^2+(y+3)^2-0.02)((x+3)^2+(y-7)^2-0.02)=0 [-12.53, 12.78, -4.8, 7.86]}