Let A(1,4) and B(1,-5) be two points.

Question

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 $(PA)^2+(PB)^2$ has maximum value, then the points $P$, $A$ and $B$ lie on

Accepted Answer

P be a point on ${(x - 1)^2} + {(y - 1)^2} = 1$

so $$P(1 + \cos \theta ,1 + \sin \theta )$$

A(1, 4), B(1, -5)

${(PA)^2} + {(PB)^2}$

$$ = {(\cos \theta )^2} + {(\sin \theta  - 3)^2} + {(\cos \theta )^2} + {(\sin \theta  + 6)^2}$$

$$ = 47 + 6\sin \theta $$

It is maximum if $$\sin \theta  = 1$$

When $$  \sin \theta  = 1,\cos \theta  = 0$$

So P(1, 2), A(1, 4), B(1, -5)

P, A, B are collinear points.

Solutions · 1