Question

A circle has a center that falls on the line `y = 11/7x +8` and passes through `( 9 ,1 )` and `(8 ,7 )`. What is the equation of the circle?

Answer 1

`(x+3.8559)^2+(y-1.9407)^2=(12.8903)^2`

Explanation:

We will make two equations with two variables, where `(x,y)` is the center of the circle.

The first equation is the given line, which passes through the center of the circle. Isolate `x` so it's more convenient to substitute:
`y=11/7x+8`
`y-8=11/7x`
`x=7/11y-56/11`

We know that the points `(9,1)` and `(8,7)` are an equal distance away from the center, so use the distance formula (direct variant of Pythagorean theorem) to determine the center and radius:

`"Distance "=sqrt((y_2-y_1)^2+(x_2-x_1)^2)`

`sqrt[(11/7x+8-color(blue)(7))^2+(7/11y-56/11-color(blue)(8))^2]=sqrt[(11/7x+8-color(blue)(1))^2+(7/11y-56/11-color(blue)(9))^2]`

Substitute `y=11/7x+8` into the equation above:

`sqrt[(11/7x+8-color(blue)(7))^2+(7/11 color(red)((11/7x+8)) -56/11-color(blue)(8))^2]=sqrt[(11/7x+8-color(blue)(1))^2+(7/11 color(red)((11/7x+8)) -56/11-color(blue)(9))^2]`

`sqrt[(11/7x+1)^2+(x-8)^2]=sqrt[(11/7x+7)^2+(x-9)^2]`

The exact answers can be found by squaring each side, then expanding the polynomials to get a quadratic equation, but only one of the values for `x` will work when substituted back in.

(Using a graphing calculator)
`color(green)(x approx -3.8559)`

`color(green)(y approx 1.9407`

`color(green)("Radius"approx 12.8903` (distance from center to either one of the points)

Now plug these in to the standard form for the equation of a circle to get:
`(x+3.8559)^2+(y-1.9407)^2=(12.8903)^2`

Answer 2

`(x+455/118)^2+(y-229/118)^2=(9+455/118)^2+(1-229/118)^2`

Explanation:

Centre falls on `y=11/7x+8`.

as points `(9,1)` and `(8,7)` also fall on circle, centre falls on perpendicular bisector of segment joining `(9,1)` and `(8,7)`.

Observe that the slope of this segment is `(7-1)/(8-9)=-6`. Hence slope of peependicular bisector wiuld be `1/6`.

As their midpoint point is `((9+8)/2,(7+1)/2)` i.e.`(17/2,4)`, equation of perpendicular bisector is `y=1/6(x-17/2)+4=1/6x-17/12+4=1/6x+31/12`

and centre is point of intersection of `y=1/6x+31/12` and `y=11/7x+8`

i.e. `11/7x+8=1/6x+31/12` or `(11/7-1/6)x=31/12-8`

or `59/42x=-65/12` i.e. `x=-65/12xx42/59=-1365/354=-455/118`

and `y=1/6xx(-455/118)+31/12=-455/708+31/12=(-455+1829)/708=1374/708=229/118`

and centre is `(-455/118,229/118)` and radius is its distance from say `(9,1)` i.e.

`sqrt((9+455/118)^2+(1-229/118)^2)`

and equation of circle is

`(x+455/118)^2+(y-229/118)^2=(9+455/118)^2+(1-229/118)^2`

graph{((x+455/118)^2+(y-229/118)^2-(9+455/118)^2+(1-229/118)^2)(y-11/7x-8)((x-9)^2+(y-1)^2-0.04)((x-8)^2+(y-7)^2-0.04)(y-1/6x-31/12)=0 [-20.92, 19.08, -8.88, 11.12]}