In the question,
If $A$ and $B$ are points in the plane such that $\frac{PA}{PB} = k$ (constant) for all $P$ on a given circle, then the value of $k$ cannot be equal to ......?
How is the locus of $P$ the perpendicular bisector of $AB$ and further how does this imply that $k$ cannot be equal to $1$?
I couldn't get my head around this one. Assumed $A, B$ and $P$ in general coordinates but it all became a mess so I had to see the solution. Please help. Thanks.
The locus of points $P$ satisfying $\frac{AP}{BP} = k$ is the perpendicular bisector of $\overline{AB}$ only if $k=1$. Otherwise the locus is a circle. Since you are given that the locus is a circle, it must be that $k\neq1$.
To see why the locus is a circle if $k\neq1$, it's probably easiest not to use coordinates. The key is Proposition VI.3 of Euclid's Elements, which can be restated as follows. Let $Q$ be the point between $A$ and $B$ satisfying $\frac{AQ}{BQ}=k$. (For the sake of argument, we can assume that $k>1$; if we were to have $k<1$, then we could just exchange the roles of $A$ and $B$.) Then Prop. VI.3 states that, for any point $P$, the ratio $\frac{AP}{BP}$ is equal to $k$ if and only if $\overrightarrow{PQ}$ bisects $\angle APB$.
What's more, if $R$ is the point on $\overrightarrow{AB}$ on the opposite side of $B$ so that $\frac{AR}{BR}=k$, then an argument very similar to Euclid's in VI.3 can be used to prove that, for any point $P$, the ratio $\frac{AP}{BP}$ is equal to $k$ if and only if $\overrightarrow{PR}$ bisects the exterior angle at $P$. (Prove it.) The points $Q$ and $R$ are said to divide the segment $\overline{AB}$ harmonically in the ratio $k$.
Now, if $\overrightarrow{PQ}$ bisects the interior angle at $P$, and $\overrightarrow{PR}$ bisects the exterior angle at $P$, then $\angle QPR$ must be a right angle. That means that $P$ must lie on the circle with diameter $\overline{QR}$. (See Proposition III.31.)
This provides the solution to the moving-ship problem. Suppose two ships leave islands at $A$ and $B$ at the same moment, traveling in straight-line paths at constant speed, with the ship from $A$ moving at $k$ times the speed of the ship from $B$. Where could they possibly meet?
If they sail toward one another, they will meet at $Q$. If they both sail in the direction of $\overrightarrow{AB}$, then the ship from $A$ will overtake the ship from $B$ at $R$. More generally, if they each sail in the right direction, then they could meet somewhere on the circle with diameter $\overline{QR}$.
Perhaps it would help to experiment with various values of $k$, for (say) $A=(0,0)$ and $B=(12,0)$. Pick a $Q$ between the midpoint and $B$, use that to get $k$, and go from there.