$(x-1)(y-2)=5$ and $(x-1)^2+(y+2)^2=r^2$ intersect at four points $A,B,C,D$. Centroid of $\Delta ABC$ lies on $y=3x-4$, then the locus of $D$ $(x-1)(y-2)=5$ and $(x-1)^2+(y+2)^2=r^2$ intersect at four points $A,B,C,D$. If centroid of $\Delta ABC$ lies on $y=3x-4$, then what is the locus of $D$?
I did try a couple of things, but I honestly have no idea how to approach this. I checked if $(x-1)(y-2)=5$ represented a pair of straight lines, but it doesn't.
I also tried substituting the value of $(x-1)$ from the first equation in the equation of the circle, but the only useful result I got was that the sum of y-coordinates is zero since the equation after substitution does not have a $y^3$ term, only $y^4$, $y^2$, $y$ and constant terms.
Apart from that, I have no other ideas. 

Point D is the intersection of line $y=3x$ (parallel to $y=3x-4$ and through origin) and curves: $(x-1)(3x-2)=5$ and $(x-1)^2+(y)^2=r^2$ .
$\begin{cases} y=3x \\(x-1)(y-2)=5 \\ (x-1)^2+(y+2)^2=r^2 \end{cases}$
$(x-1)(3x-2)=5$
$3x^{2}-5x-3=0$
$\begin{array}{} x_{D}=\frac{5+\sqrt{61}}{6} & y_{D}=\frac{5+\sqrt{61}}{2}  \end{array}$
solving numerically
$(x_{D}-1)^2+(y_{D}+2)^2=r^2$
$r^2=\frac{335+\sqrt{97600}}{9}≈71.93444300402957$
$\left\{ \begin{array}{} (x-1)(y-2)=5 & ⇒ y=\frac{2x+3}{x-1} \\ (x-1)^2+(y+2)^2=r^2  \end{array}  \right\}$
$sol=\left( \begin{array}{left } x_{A}=-6.788336312 & y_{A}=1.358014369 \\ x_{B}=.599099974 & y_{B}=-10.471937333 \\ x_{C}=8.054194725 & y_{C}=2.708798126 \\ x_{D}=2.135041613 & y_{D}=6.405124838 \end{array}   \right)$
checking if the centroid of triangle ABC belongs to the line $y = 3x-4$
$\begin{array}{} \text{Centroid (G)} & x_{G}=\frac{x_{A}+x_{B}+x_{C}}{3} &  y_{G}=\frac{y_{A}+y_{B}+y_{C}}{3} \end{array}$
$G=(0.621652796,-2.135041613)$
$3·x_{G}-4=y_{G}$

