$(x-1)(y-2) = 5$ 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: When a rectangular hyperbola (x-1)(y-2)=5 and circle (x-1)²+(y+2)²=r² intersect at four points, use Vieta's relations on the quartic obtained by substitution to find that the sum of x-coordinates equals 4-x₄ and sum of y-coordinates equals y₄, then apply the centroid condition on the first three points to locate the fourth point D.
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