Let $$A(1, 4)$$ and $$B(1, -5)$$ be two points. Let $$P$$ be a point on the circle $$(x-1)^2 + (y-1)^2 = 1$$, such that $$(PA)^2 + (PB)^2$$ have maximum value, then the points $$P$$, $$A$$ and $$B$$ lie on

Let $$P$$ be a point on the circle $$(x - 1)^2 + (y - 1)^2 = 1$$. We can parameterize $$P = (1 + \cos\theta,\; 1 + \sin\theta)$$.

Computing $$PA^2$$ where $$A = (1, 4)$$: $$PA^2 = \cos^2\theta + (\sin\theta - 3)^2 = \cos^2\theta + \sin^2\theta - 6\sin\theta + 9 = 10 - 6\sin\theta$$.

Computing $$PB^2$$ where $$B = (1, -5)$$: $$PB^2 = \cos^2\theta + (\sin\theta + 6)^2 = 1 + 12\sin\theta + 36 = 37 + 12\sin\theta$$.

Therefore $$PA^2 + PB^2 = 47 + 6\sin\theta$$. This is maximized when $$\sin\theta = 1$$, i.e., $$\theta = \dfrac{\pi}{2}$$, giving $$P = (1, 2)$$.

Now we check: $$P = (1, 2)$$, $$A = (1, 4)$$, $$B = (1, -5)$$. All three points have the same $$x$$-coordinate equal to 1, so they are collinear and lie on the vertical line $$x = 1$$.

Hence the points $$P$$, $$A$$, and $$B$$ lie on a straight line.