Question

A circle has a center that falls on the line `y = 2/7x +3` and passes through `( 1 ,4 )` and `(6 ,1 )`. What is the equation of the circle?

Answer 1

`(x-133/29)^2+(y-125/29)^2 = 10897/29^2`

Or if you prefer:

`(29x-133)^2+(29y-125)^2 = 10897`

Explanation:

Since the circle passes through `(1, 4)` and `(6, 1)`, its centre must also lie on the perpendicular line through the midpoint of the chord joining these two points.

Let `A = (1, 4)` and `B = (6, 1)`.

The midpoint of `AB` is `((1+6)/2, (4+1)/2) = (7/2, 5/2)`

The slope of `AB` is `(1-4)/(6-1) = -3/5`.

So the slope of any line perpendicular to `AB` is `5/3`

So the equation of the line through the midpoint of `AB` and perpendicular to it can be written in point slope form as:

`(y - 5/2) = 5/3(x-7/2)`

Add `5/2` to both sides and simplify to find the equation in slope intercept form:

`y = 5/3x - 10/3`

Find the intercept of this line and the given line by equating the values of `y`:

`5/3x - 10/3 = y = 2/7x + 3`

Multiply both ends by `21` to get:

`35x - 70 = 6x + 63`

Add `70` to both sides to get:

`35x = 6x + 133`

Subtract `6x` from both sides to get:

`29x = 133`

Hence `x = 133/29`

So:

`y = 5/3x - 10/3 = 5/3*133/29 - 10/3 = (665-290)/57 = 375/57 = 125/29`

So the equation of our circle can be written in the form:

`(x-133/29)^2+(y-125/29)^2 = r^2`

Since the circle passes through `(1, 4)` this point must satisfy the equation. So:

`r^2 = (1-133/29)^2+(4-125/29)^2`

`=(-104/29)^2+(-9/29)^2`

`=(104^2+9^2)/29^2`

`=(10816+81)/29^2`

`=10897/29^2`

So the equation of the circle may be written:

`(x-133/29)^2+(y-125/29)^2 = 10897/29^2`

graph{((x-133/29)^2+(y-125/29)^2-10897/29^2)((x-1)^2+(y-4)^2-0.02)((x-6)^2+(y-1)^2-0.02)(y - 2/7x-3)((x-133/29)^2+(y-125/29)^2-0.04)=0 [-6.58, 13.42, -1, 9]}