Question

How do you write an equation of a circle with diameter that has endpoints (-7,-2), and (-15,6)?

Answer 1

The equation of the circle is `(x+11)^2+(y-2)^2=32`

Explanation:

The points on the diameter are `A=(-7,-2)` and `B=(-15,6)`

The midpoint of `AB` is the center of the circle

`O=(((-7)+(-15))/2, ((-2)+(6))/2)=(-11,2)`

The radius of the circle is

`r^2=((-7+11)^2+(-2-2)^2)=(4^2+4^2)=32`

The equation of the circle is

`(x+11)^2+(y-2)^2=r^2=32`

graph{(x+11)^2+(y-2)^2=32 [-23.04, 2.27, -3.43, 9.24]}