Locus with constant arc length restraint Please help finding locus of point $P$  through $(0,1)$ moving in the plane so that  arc starting on $x-$ axis with a vertical tangent as shown  has the same arc length $=\pi/2,$ upto $P$. I.e., $AP= A_1P_1$ in answered sketch.
EDIT1:
Picture of what I had in mind at question time and Hyperbolic spiral answer:
EDIT2:
Actually at that time I was imagining profile of the edges of a bulky book.

 A: You may set up a parametric equation as follows:
$$(x(t),y(t)) = r(t)(\cos{\alpha(t)},\sin{\alpha(t)})$$
where


*

*$r(t) = y(t) = 1-t$ with $t \in [0,1)$

*$\alpha(0) = \frac{\pi}{2}$

*$\alpha(t) = \alpha(0) - \beta(t)$ where $r(t)\cdot\left( \frac{\pi}{2} + \beta(t) \right) = \frac{\pi}{2}$ as the arc length of the smaller circle needs to be $\frac{\pi}{2}$ (here we see already that the curve must spiral with decreasing $r$)


From the last condition you get 
$$\beta(t) = \frac{\pi}{2}\left( \frac{1-r(t)}{r(t)} \right) =  \frac{\pi}{2}\left( \frac{t}{1-t} \right)$$
So, you get your $\alpha$:
$$\alpha(t) = \alpha(0) - \beta(t) =  \frac{\pi}{2} -  \frac{\pi}{2}\left( \frac{t}{1-t} \right) =  \frac{\pi}{2}\left( \frac{1-2t}{1-t} \right)$$
All together:
$$(x(t),y(t)) = (1-t)(\cos{\frac{\pi}{2}\left( \frac{1-2t}{1-t} \right)},\sin{\frac{\pi}{2}\left( \frac{1-2t}{1-t} \right)})$$
And that's how it looks like:

p.s.:
Unfortunately the one who asked did not specify that he/she wanted an answer in polar coordinates.
The third condition I mentioned above means:
$$r(\theta)\cdot\left( \frac{\pi}{2} +\beta \right) =  \frac{\pi}{2}$$
where $\beta = \frac{\pi}{2} - \theta$, if $\theta$ is measured anticlockwise from the positive $x$-axis.
So, you get 
$$r(\theta) = \frac{\pi}{2(\pi - \theta)} \mbox{ with } \theta_0 = \frac{\pi}{2}$$
