How do you find the equation of the circle in general form containing the points (-3,-5), (1,3), and (5,-1)?

1 Answer
Dec 29, 2015

Answer:

Use an alternative, slightly more geometric method (not necessarily shorter) to find:

#(x-1/3)^2+(y+5/3)^2 = ((10 sqrt(2))/3)^2#

Explanation:

The centre of the circle can also be found as the intersection of two lines using linear equations as follows:

Let #A=(-3, -5)#, #B=(1, 3)# and #C=(5, -1)#

The midpoint of #AB# is #((-3+1)/2, (-5+3)/2) = (-1, -1)#

The slope of #AB# is #(3-(-5))/(1-(-3)) = 8/4 = 2#

So any perpendicular to #AB# has slope #-1/2#

The perpendicular through the midpoint of #AB# can be written:

#color(blue)(y+1 = -1/2 (x+1))#

The points on this line are equidistant from #A# and #B#.

The midpoint of #BC# is #((1+5)/2, (3-1)/2) = (3, 1)#

The slope of #BC# is #(-1-3)/(5-1) = (-4)/4 = -1#

So any perpendicular to #BC# has slope #-(1/-1) = 1#

The perpendicular through the midpoint of #BC# can be written:

#color(blue)(y-1 = x-3)#

The points on this line are equidistant from #B# and #C#.

Subtract the two equations of the perpendicular lines to find:

#2 = -1/2(x+1) - (x-3) =-3/2x+5/2#

Multiply both sides by #2# to get:

#4 = -3x+5#

Subtract #5# from both sides:

#-1 = -3x#

Divide both sides by #-3#:

#x = 1/3#

Then substitute this into the second equation to find #y = -5/3#

So the circle has equation #(x-1/3)^2+(y+5/3)^2 = r^2# for some radius #r#.

Substitute point #B# into this equation to find:

#r^2 = (1-1/3)^2 + (3+5/3)^2 =4/9+ 196/9 = 200/9#

So #r = sqrt(200/9) = (10 sqrt(2))/3#

So our equation in standard form is:

#(x-1/3)^2+(y+5/3)^2 = ((10 sqrt(2))/3)^2#