Question

A circle has a center that falls on the line `y = 7/4x +4` and passes through `( 3 ,7 )` and `(7 ,5 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x-32)^2+(y-60)^2=3650`

Explanation:

Let the center of the circle be `(a,b)` and radius r
Then the equation of the circle is
`(x-a)^2+(y-b)^2=r^2`

Then, `(3-a)^2+(7-b)^2=r^2`
and `(7-a)^2+(5-b)^2=r^2`

Equqting the equations
`(3-a)^2+(7-b)^2=(7-a)^2+(5-b)^2`
developing,
`9-6a+a^2+49-14b+b^2=49-14a+a^2+25-10b+b^2`

`9-6a-14b=25-14a-10b`
`8a-4b=16`
`2a-b=4` this is equation 1
Also, `y=(7x)/4+4`
So `b=(7a)/4+4`
Solving for a and be, we get `a=32` and `b=60`
We calculate the radius `r^2=(7-32)^2+(5-60)^2`

`r^2=25^2+55^2` `=>` `r=60.41`

Equation of circle is
`(x-32)^2+(y-60)^2=3650`