How do you write the equation for a circle where the points passes through the points (1,1), (-2, 2), and (-5,1)?

1 Answer
Nov 13, 2016

Answer:

Please see the explanation for the steps leading to:

#(x - -2)^2 + (y - -3)^2 = 5^2#

Explanation:

Use the standard form for the equation of a circle:

#(x - h)^2 + (y - k)^2 = r^2#

And the points, #(1,1), (-2,2), and (-5,1)#

To write 3 equations:

#(1 - h)^2 + (1 - k)^2 = r^2" [1]"#
#(-2 - h)^2 + (2 - k)^2 = r^2" [2]"#
#(-5 - h)^2 + (1 - k)^2 = r^2" [3]"#

Temporarily eliminate the variable r by setting the left side of equation [1] equal to the left side of equation [2] and the same for equation [1] and equation [3]:

#(1 - h)^2 + (1 - k)^2 = (-2 - h)^2 + (2 - k)^2#
#(1 - h)^2 + (1 - k)^2 = (-5 - h)^2 + (1 - k)^2" [4]"#

Please notice that equation [4] has #(1 - k)^2# on both sides so we can subtract this form both sides:

#(1 - h)^2 + (1 - k)^2 = (-2 - h)^2 + (2 - k)^2#
#(1 - h)^2 = (-5 - h)^2#

Use the pattern #(a - b)^2 = a^2 - 2ab + b^2# to expand all of the squares:

#1 - 2h + h^2 + 1 - 2k + k^2 = 4 + 4h + h^2 + 4 - 4k + k^2#
#1 - 2h + h^2 = 25 + 10h + h^2#

The #h^2 and k^2# terms cancel:

#1 - 2h + 1 - 2k = 4 + 4h + 4 - 4k" [5]"#
#1 - 2h = 25 + 10h" [6]"#

Use equation 6 to solve for h:

#-12h = 24#

#h = -2#

Combine like tems of equation [5] with k terms on the left and constant and h terms on the right:

#2k = 6h + 6#

#k = 3h + 3#

Substitute -2 for h:

#k = 3(-2) + 3#

#k = - 3#

Substitute the values for h and k into one of the first three equations (I will use equation [2]):

#(-2 - -2)^2 + (2 - -3)^2 = r^2" [2]"#

#(0)^2 + (5)^2 = r^2#

Solve for r:

#r = 5#

The equation of the circle is:

#(x - -2)^2 + (y - -3)^2 = 5^2#