Question

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

Answer 1

The equation of the circle is `(x+40/13)^2+(y+18/13)^2=10.51^2`

Explanation:

Let `(a,b)` be the center of the circle
As the center lies on the line, we have `b=(7a)/4+4`
Then `(x-a)^2+(y-b)^2=r^2`, and `r` is the radius of the circle
So, `(3-a)^2+(7-b)^2=r^2`
and `(7-a)^2+(1-b)^2=r^2`
`:.` `(3-a)^2+(7-b)^2=(7-a)^2+(1-b)^2`
`9-6a+a^2+49-14b+b^2=49-14a+a^2+1-2b+b^2`
`58-6a-14b=50-14a-2b`
`12b=8a+8` `=>` `3b=2a+2`
Solving for `(a,b)` with the two equations
We find `a=-40/13` and `b=-18/13`

Then `r=sqrt((7+(40/13))^2+(1+(18/13))^2)`
`r=10.51`
Having `r`, `a`, and `b`, we write the equation of the circle