Question

A circle has a center that falls on the line `y = 8/7x +6` and passes through `( 2 ,1 )` and `(3 ,6 )`. What is the equation of the circle?

Answer 1

`(x - -70/47)^2 + (y - 202/47)^2 = (sqrt(50921)/47)^2`

Explanation:

The standard equation of a circle is

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

where `(h, k)` is the center point and `r` is the radius.
Therefore, the given equation of a the line at the center point is:

`k = 8/7h + 6` [1]

The standard equation evaluated at the two given points gives us the two equations:

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

There are 3 equations and 3 unknown values.

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

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

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

`4 - 4h + h^2 + 1 - 2k + k^2 = 9- 6h + h^2 + 36 - 12k + k^2`

Cancel the square terms:

`4 - 4h + cancel(h^2) + 1 - 2k + cancel(k^2) = 9- 6h + cancel(h^2) + 36 - 12k + cancel(k^2)`

Combine the constant terms into a single term on the right:

`-4h- 2k = -6h + 40 - 12k`

Combine the h terms into a single term on the right:

`-2k = -2h + 40 - 12k`

Combine the k terms into a single term on the left:

`10k = -2h + 40`

Divide both sides by 10:

`k = -1/5h + 4` [4]

Subtract equation [4] from equation [1]:

`k -k = 8/7h + 1/5h + 6 - 4`

`0 = 8/7h + 1/5h + 2`

`0 = 47/35h + 2`

`h = -70/47`

Substitute this value for h into equation [1]:

`k = 8/7(-70/47) + 6`

`k = 8/7(-70/47) + 6`

`k = 202/47`

Substitute these value for h and k into equation [2]:

`(2 - -70/47)^2 + (1 - 202/47)^2 = r^2`

`r^2 = 50921/47^2`

`r = sqrt(50921)/47`

Check with equation [3]:

`(3 - -70/47)^2 + (6 - 202/47)^2 = r^2`

`50921/47^2 = r^2`

This checks