Idk either
I found that the centre of the circle is \((a,\frac{9a-16}{3})\) where a is int the set ot reals
and the radius sqared iis \(r^2= 10a^2-40a+\frac{370}{9}\)
so the equation of the circle is
\((x-a)^2+(y-\frac{9a-16}{3})^2= 10a^2-40a+\frac{370}{9}\)
I know that is right becasue I have graphed it but when i take it further I get in a mess. :/