Question

(-2,1) and (4,1) are the endpoints of one chord of the circle, and (-2,-3) and (4,-3) are the endpoints of another chord of the circle. What is the center, radius, and equation?

Answer 1

Center is `(1,-1)`, radius is `sqrt13` and equation of circle is `x^2+y^2-2x+2y-11=0`

Explanation:

As the ordinates of the endpoints of the chord joining `(-2,1)` and `(4,1)` are equal, its perpendicular bisector is parallel to `y`-axis and its length is `4-(-2)=6`. Note midpoint of chord is `((-2+4)/2,(1+1)/2)` or `(1,1)`.

Similarly, as the ordinates of the endpoints of the chord joining `(-2,-3)` and `(4,-3)` are equal, its perpendicular bisector is also parallel to `y`-axis and its length too is `4-(-2)=6`. Note midpoint of chord is `((-2+4)/2,(-3-3)/2)` or `(1,-3)`

and hence two chords are equal and parallel and hence center is midpoint of the segment joining their midpoints i.e. `((1+1)/2,(1-3)/2)` or `(1,-1)`.`

Hence, radius is distance between `(1,-1)` and say `(4,1)` i.e.

`sqrt((4-1)^2+(1+1)^2)=sqrt13` and equation of circle is

`(x-1)^2+(y+1)^2=13` i.e. `x^2+y^2-2x+2y-11=0`

graph{x^2+y^2-2x+2y-11=0 [-8.92, 11.08, -6.32, 3.68]}