Question

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

Answer 1

The equation of a circle with center `(x_o,y_o)` and radius `R` is

`(x-x_o)^2+(y-y_o)^2=R^2`

Hence the center belongs to the line we have that

`y_o=1/7*x_o +4`

Because the circle passes from point `(7,8)` and `(3,1)` we have also

`(7-x_o)^2+(8-y_o)^2=R^2`

and

`(3-x_o)^2+(1-y_o)^2=R^2`

Hence the RHS of both equations is the same (`=R^2`)
we have

`(7-x_o)^2+(8-y_o)^2=(3-x_o)^2+(1-y_o)^2`

#7^2-14*x_o + (x_o)^2+8^2-16*y_o + (y_o)^2= 3^2-6*x_o + (x_o)^2+1-2*y_o + (y_o)^2#

`49+64-14*x_o-16*y_o-9+6*x_o-1+2*y_o=0`

`-8*x_o-14*y_o + 103=0`

`8*x_o + 14*y_o - 103=0`

Because `y_o=1/7*x_o +4` and `8*x_o + 14*y_o - 103=0`

solving the system of these two equation yields the center of the

circle

`(x_o,y_o)=(47/10,327/70)`

and the radius is

`(3-47/10)^2+(1-327/70)^2=R^2`

`R=sqrt(8021/490)`

Finally the equation of the circle is

`(x-47/10)^2+(y-327/70)^2=8021/490`