JEE Main 2021 (Online) 26th February Evening ShiftQuestion 1
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 :
Circle
Solution
Let M=(1,-1/2) be midpoint of AB. $PA^2+PB^2=2PM^2+AB^2/2$, maximum when P is farthest from M on the circle. Centre $(1,1)$ lies on $x=1$, so the farthest point is $P=(1,2)$. A, B, P all lie on $x=1$: a straight line. Option (D).