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

1 Answer
Feb 9, 2016

Answer:

#(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]}