(x-1)(y-2)=5 and `(x-1)^2+(y+2)^2=r^2` intersect at four points A, B, C, D and if centroid of `triangle ABC` lies on line `y = 3x-4` , then locus of D is

Explanation

Given hyperbola is
`(x-1)(y-2)=5" (1)"`
and circle is
`(x-1)^(2)+(y+2)^(2)=r^(2)" (2)"`
These curves intersect at four points A, B, C and D.
We know curves intersect at four points is midpoint of centres of curves.
`(sum_(i=1)^4x_(i))/(4)=(1+1)/(2)=1 and (sum_(i-1)^(4)y_(1))/(4)=(2-2)/(2)=0,`
where `(x_(i),y_(i),i=1,2,3,4` are points A, B, C and D.
`therefore" "sum_(i=1)^(4)x_(1)=4 and sum_(i=1)^(4)y_(i)=0`
`therefore" "x_(1)+x_(2)+x_(3)=4-x_(4) and y_(1)+y_(2)+y_(3)=-y_(4)`
So, centroid of triangle ABC is
`((x_(1)+x_(2)+x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3))-=((4-x_(4))/(3),(-y_(4))/(3))`
Now, given that centroid lies on the line `y = 3x-4`.
`therefore" "(-y_(y))/(3)=3((4-x_(4))/(3))-4`
`therefore" "y_(4)=3x_(4)`
Therefore, y=3x, which is locus of point D.