Question

A circle has a center that falls on the line `y = 3/2x +1` and passes through `(1 ,2 )` and `(6 ,1 )`. What is the equation of the circle?

Answer 1

The equation of the circle is:
`(x - 34/7)^2 + (y - 58/7)^2 = (sqrt2665/7)^2`

Explanation:

The equation of a circle is:

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

where `(h,k)` is the center and r is the radius.

Substitute the point (h,k) into the equation for the line on which the center lies:

`k = 3/2h + 1` `" [1]"`

We can use the two given points and the equation of a circle to write two more equations:

`(1 - h)^2 + (2 - k)^2 = r^2` `" [2]"`

`(6 - h)^2 + (1 - k)^2 = r^2` `" [3]"`

We can, temporarily, eliminate the variable r by setting the left side of equation [2] equal to the left side of equation [3]:

`(1 - h)^2 + (2 - k)^2 = (6 - h)^2 + (1 - k)^2`

Use the pattern `(a - b)^2 = a^2 - 2ab + b^2`, to expand the squares:

`1 - 2h + h^2 + 4 - 4k + k^2 = 36 - 12h + h^2 + 1 - 2k + k^2`

The square terms are the same on both sides of the equation, therefore, they cancel:

`1 - 2h + 4 - 4k = 36 - 12h + 1 - 2k`

Collect the 4 constant terms into a single term on the right:

`-2h - 4k = 32 - 12h - 2k`

Collect the 2 h terms into a single term on the right:

`-4k = 32 - 10h - 2k`

Collect the 2 k terms into a single term on the left:

`-2k = 32 - 10h`

Divide both sides by -2:

`k = 5h - 16` `" [4]"`

Subtract equation [4] from equation [1]:

`k - k = 3/2h - 5h + 1 + 16`

`0 = -7/2h + 17`

`h = 34/7`

Substitute 34/7 for h in equation [1]:

`k = 3/2(34/7) + 1`

`k = 58/7`

Substitute the values for h and k into equation [2] or [3] (I will use equation [2]):

`(1 - 34/7)^2 + (2 - 58/7)^2 = r^2`

`(-27/7)^2 + (-44/7)^2 = r^2`

`r^2 = 2665/7^2`

`r = sqrt(2665)/7`

The equation of the circle is:
`(x - 34/7)^2 + (y - 58/7)^2 = (sqrt2665/7)^2`