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  4. 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 :

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Q. 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 :

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Solution:

$P$ be a point on $( x -1)^{2}+( y -1)^{2}=1$
so $P (1+\cos \theta, 1+\sin \theta)$
$\begin{array}{ll} A (1,4) & B (1,-5)\end{array}$
$( PA )^{2}+( PB )^{2}$
$=(\cos \theta)^{2}+(\sin \theta-3)^{2}+({cos} \theta)^{2}+(\sin \theta+6)^{2}$
$=47+6 \sin \theta$
is maximum if $\sin \theta=1$
$\Rightarrow \sin \theta=1, \cos \theta=0$
$P (1,1) A (1,4) B (1,-5)$
$P , A , B$ are collinear points.