If $$P(6, 1)$$ be the orthocentre of the triangle whose vertices are $$A(5, -2)$$, $$B(8, 3)$$ and $$C(h, k)$$, then the point $$C$$ lies on the circle:

Orthocentre P(6,1), A(5,-2), B(8,3). AP⊥BC: slope AP = (1+2)/(6-5) = 3. Slope BC = -1/3.

BP⊥AC: slope BP = (1-3)/(6-8) = 1. Slope AC = -1.

BC: passes through B(8,3) slope -1/3: y-3 = -1/3(x-8) → x+3y = 17.

AC: passes through A(5,-2) slope -1: y+2 = -(x-5) → x+y = 3.

C is intersection: x+3y=17, x+y=3 → 2y=14, y=7, x=-4. C(-4,7).

C lies on circle: x²+y² = 16+49 = 65.

The correct answer is Option (3): x²+y²-65=0.