(-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?

1 Answer
Mar 28, 2017

Answer:

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]}