$(x-1)(y-2) = 5$ 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:
Step-by-Step Solution
Key Concept: For four intersection points of a rectangular hyperbola (x-1)(y-2)=5 and a circle, use the property that the sum of coordinates of all four points relates to the algebraic structure of the curves. When the centroid of three points A, B, C lies on y=3x-4, the fourth point D's coordinates satisfy the same linear relation due to the symmetry properties of the intersection configuration.
For four intersection points $(x_i, y_i)$, the centroid coordinates satisfy $\frac{\sum x_i}{4} = \frac{1 + 1}{2} = 1$ and $\frac{\sum y_i}{4} = 0$. Computing $\sum x_i = 4 - x_4$ and $\sum y_i = y_4$, the centroid lies on $y = 3x - 4$, yielding $y_4 = 3x_4$.
Correct Answer: 1