Let $\mathrm{A}(1,4)$ and $\mathrm{B}(1,-5)$ be two points. Let $\mathrm{P}$ be a point on the circle $(x-1)^{2}+(y-1)^{2}=1$ such that $(\mathrm{PA})^{2}+(\mathrm{PB})^{2}$ have maximum value, then the points $P, A$ and $B$ lie on :
Correct Option: 1
$P$ be a point on $(x-1)^{2}+(y-1)^{2}=1$
so $\mathrm{P}(1+\cos \theta, 1+\sin \theta)$
$\mathrm{A}(1,4) \quad \mathrm{B}(1,-5)$
$(\mathrm{PA})^{2}+(\mathrm{PB})^{2}$
$=(\cos \theta)^{2}+(\sin \theta-3)^{2}+(\operatorname{cso} \theta)^{2}+(\sin \theta+6)^{2}$
$=47+6 \sin \theta$
is maximum if $\sin \theta=1$
$\Rightarrow \sin \theta=1, \cos \theta=0$
$\mathrm{P}(1,1) \mathrm{A}(1,4) \mathrm{B}(1,-5)$
$\mathrm{P}, \mathrm{A}, \mathrm{B}$ are collinear points.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.