Question

A circle has a center that falls on the line `y = 7/9x +5` and passes through `( 7 ,3 )` and `(5 ,1 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x-27/16)^2+(y-101/16)^2=39.2`

Explanation:

Let the center of the circle be `(a,b)`
Then the equation of the circle is `(x-a)^2+(y-b)^2=r^2`
As the circles passes through `(7,3)` and `(5,1)`
Then substituting those points in the equation of the circle

`(7-a)^2+(3-b)^2=r^2`
`(5-a)^2+(1-b)^2=r^2`

so `(7-a)^2+(3-b)^2=(5-a)^2+(1-b)^2`
Developing both sides
`49-14a+a^2+9-6b+b^2=25-10a+a^2+1-2b+b^2`
simplifying
`58-14a-6b=26-10a-2b`
`4a+4b=32`
`a+b=8` this is equation 1
Putting `(a,b)` in the equation of the line
`b=7/9a+5` this is equation 2
Solving for a and b
we obtain `(27/16,101/16)` as the center of the circle
The radius `r=sqrt (85^2+63^2)/16=6.26`
The equation of the circle is `(x-27/16)^2+(y-101/16)^2=39.2`