Question

A circle has a center that falls on the line `y = 8/9x +4` and passes through `( 4 ,1 )` and `(3 ,7 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x+63/78)^2+(y-128/39)^2=172309/6084`

Explanation:

So the center C of the circle falls on the line `y=(8x+36)/9`
If we find the equation of other line in which a radius of the circle stands, this two lines will meet at the circle's center because the circle's radii converge to its center.

Calling
`A(4,1)`
`B(3,7)`
The segment AB is a chord of the circle that is divided in the precise middle by a radius that is perpendicular to this chord.

`M(3.5, 4)`
Slope of the line in which segment AB stands and of a line perpendicular to it
`k=(Delta y)/(Delta x)=(7-1)/(3-4)=6/(-1)=-6` => `p=-1/k=1/6`

Other equation in which a radius stands:
`MC -> y-4=(1/6)(x-3.5)` => `y=(x-3.5)/6+4` => `y=(x+20.5)/6`

Finding the intersection of the two lines:
`(8x+36)/9=(x+20.5)/6` => `48x+216=9x+184.5` => `39x=-31.5` => `x=-63/78`
`-> y=(-63/78+41/2)/6=(-63+1599)/468=1536/468` => `y=128/39`
=> `C(-63/78,128/39)`

Finding the radius
`r=sqrt((4+63/78)^2+(1-128/39)^2)=(sqrt((312+63)^2+4*(39-128)^2))/78=sqrt(140625+31684)/78=sqrt(172309)/78~=5.414`

Then the equation of the circle is
`(x-x_c)^2+(y-y_c)=r^2`
`(x+63/78)^2+(y-128/39)^2=172309/6084`