Question

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

Answer 1

`(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]}