Question

How do you write the equation for a circle endpoints of a diameter at (3,8) and (-7,4)?

Answer 1

The equation of circle is `x^2+y^2+4x-12y+11=0`

Explanation:

The center of the circle is mid point between `(3,8)` and `(-7,4)` i.e. `((3-7)/2,(8+4)/2` or `(-2,6)`.

The radius is distance between `(-7,4)` and `(-2,6)` or `sqrt((-2+7)^2+(6-4)^2)=sqrt(25+4)=sqrt29`

Hence equation of circle of radius `sqrt29` and center `(-2,6)` is

`(x+2)^2+(y-6)^2=29` or

`x^2+4x+4+y^2-12y+36-29=0` or

`x^2+y^2+4x-12y+11=0`