Question

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

Answer 1

THe equation of the circle is `(x-11/16)^2+(y-117/16)^2=58.4`

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`
Putting the points `(7,3)` and `(5,1)` in the equation of the circle

`(7-a)^2+(3-b)^2=r^2` and `(5-a)^2+(1-b)^2=r^2`
so `(7-a)^2+(3-b)^2=(5-a)^2+(1-b)^2`
Expanding the equations

`49-14a+a^2+9-6b+b^2=25-10a+a^2+1-2b+b^2`

`49-14a+9-6b=25-10a+1-2b`

`58-14a-6b=26-10a-2b`
`32=4a+4b` `=>` `8=a+b` this is equation 1

From the equation of the line, we have `b=(7a)/9+1`
`9b=7a+9` this is equation 2
Solving for a and b , we get
`a=11/16` and `b=117/16`

Then we calculate `r`
`r^2=(7-11/16)^2+(3-117/16)^2`
`r^2=(101^2+69^2)/16`
`r=7.65`
So the equation of the circle is
`(x-11/16)^2+(y-117/16)^2=58.4`