Question

A circle has a center that falls on the line `y = 5/8x +6` and passes through `( 1 ,4 )` and `(2 ,9 )`. What is the equation of the circle?

Answer 1

`(x - 32/33)^2 + (y - 218/33)^2 = (sqrt(7397)/33)^2`

Explanation:

The standard Cartesian form of the equation of a circle is:

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

where x and y correspond to any point, `(x, y)`, on the circle, h and k correspond to the center point, `(h, k)`, and r is the radius.

Use equation [1] and the points `(1, 4) and (2,9)` to write two equations:

`(1 - h)^2 + (4 - k)^2 = r^2" [2]"`
`(2 - h)^2 + (9 - k)^2 = r^2" [3]"`

Expand the squares on the left sides of equations [2] and [3], using the pattern, `(a - b)^2 = a^2 - 2ab + b^2`:

`1 - 2h + h^2 + 16 - 8k + k^2 = r^2" [4]"`
`4 - 4h + h^2 + 81 - 18k + k^2 = r^2" [5]"`

Subtract equation [5] from equation [4]:

`1 - 2h + h^2 + 16 - 8k + k^2 - (4 - 4h + h^2 + 81 - 18k + k^2) = r^2 - r^2`

`1 - 2h + h^2 + 16 - 8k + k^2 - 4 + 4h - h^2 - 81 + 18k - k^2 = 0`

`1 - 2h + 16 - 8k - 4 + 4h - 81 + 18k = 0`

`2h + 10k = 68`

`h + 5k = 34" [6]"`

Substitute h for x and k for y into the given equation, `y = 5/8x + 6`:

`k = 5/8h + 6" [7]"`

Substitute the right side of equation [7] for k into equation [6]:

`h + 5(5/8h + 6) = 34`

Solve for h:

`h + 25/8h + 30 = 34`

`33/8h = 4`

`h = 32/33`

Substitute this value for h into equation [7]:

`k = 5/8(32/33) + 6`

`k = 218/33`

Substitute the values for h and k into equation [3]:

`(2 - 32/33)^2 + (9 - 218/33)^2 = r^2`

Solve for r:

`(66/33 - 32/33)^2 + (297/33 - 218/33)^2 = r^2`

`(34/33)^2 + (79/33)^2 = r^2`

`r^2 = 7397/33^2`

`r = sqrt(7397)/33`

Substituting the values for, h, k, and r into equation [1], we obtain the equation:

`(x - 32/33)^2 + (y - 218/33)^2 = (sqrt(7397)/33)^2`

The following is a graph of the center point, the two given points, the given line and the circle: