Question

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

Answer 1

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`