Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$  represent the sides of a triangle whose centroid is G, then the equation of the circle  which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is

The centroid $G$ belongs to the line
$\mathcal{L}:$ $y=x+1$.
Since the circle $\mathcal{C}$ must pass through
$G(2,3)$ and $D(1,1)$, it must also pass through
the point $K(4,4)$,
which is a reflection of $D$ wrt the perpendicular through $G$.
Hence, the sought circle is the circumcircle of $\triangle GDK$
with the side lengths
\begin{align} 
|GD|=|GK|&=\sqrt5
,\\
|DK|&=3\sqrt2
,
\end{align}
the area
\begin{align} 
S_{GDK}&=
\tfrac12\,|DK|\,\sqrt{|GD|^2-\tfrac14\,|DK|^2}
=\tfrac32
\end{align}
and the circumradius
\begin{align} 
R&=\frac{|GD|^2|DK|}{4S_{GDK}}
=\tfrac52\,\sqrt2
.
\end{align}
Since in this special simple case
$GO$ is the diagonal
of the $\tfrac52\times\tfrac52$ square,
the coordinates of the center of the circle $\mathcal C$
can be found as
\begin{align} 
O&=G+(\tfrac52,\, -\tfrac52)
=(\tfrac92,\, \tfrac12)
.
\end{align}
In case of general $\triangle ABC$
with the side lengths $a,b,c$,
we could find
the center $O$
using a known expression
\begin{align}
O&=
\frac{a^2\,(b^2+c^2-a^2)\,A+b^2\,(a^2+c^2-b^2)\,B+c^2\,(b^2+a^2-c^2)\,C}
{a^2\,(b^2+c^2-a^2)+b^2\,(a^2+c^2-b^2)+c^2\,(b^2+a^2-c^2)}
\\
&=
\frac{a^2\,(b^2+c^2-a^2)\,A+b^2\,(a^2+c^2-b^2)\,B+c^2\,(b^2+a^2-c^2)\,C}
{16S_{ABC}^2}
,
\end{align}
which, of course, works in this special case as well
and provides the same result.
