Question

A circle has a diameter with endpoints (-10, -6) and (-2, -4). What is the equation of the circle?

Answer 1

`(x+6)^2+(y+5)^2=17`

Explanation:

A line segment with end points `(-10,-6)` and `(-2,-4)`

• has a mid point at `(-6,-5)`
• has a length of `sqrt((-10-(-2))^2+(-6-(-4)^2))= sqrt(8^2+2^2)=sqrt(68) =2sqrt(17)`

If the line segment represents a diameter for a circle:
the circle must:

• have its center at `(color(red)(-6),color(blue)(-5))`
• have a radius of `(2sqrt(17))/2 =color(green)( sqrt(17))`

The general equation of circle with center at `(color(red)(a),color(blue)(b))` and radius `color(green)(r)` is
`color(white)("XXX")(x-color(red)(a))^2+(y-color(blue)(b))^2=color(green)(r)^2`

Substituting:
`color(white)("XXX")color(red)(a=-6)`
`color(white)("XXX")color(blue)(b=-5)`
`color(white)("XXX")color(green)(r=sqrt(17)`

We get:
`color(white)("XXX")(x-(color(red)(-6)))^2+(y-(color(blue)(-5)))^2=(color(green)(sqrt(17)))^2`
or
`color(white)("XXX")(x+6)^2+(y+5)= 17`

The graph below of this equation might help to verify that it does seem to match the given end point conditions.
graph{(x+6)^2+(y+5)^2=17 [-14.16, 5.84, -9.76, 0.235]}