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

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