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A circle is inscribed in a square $ABCD$ of side length $2$. There is a point $P$ on the circle such that $PA=a$. Is it possible to find $PB,PC,PD$ in terms of $a$?

I haven't solved a problem like this before, so I'm not sure how I should approach it. I'm looking to solve it without actually computing anything if possible.

Thanks

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If you draw a sketch you will see that the locus of points $P$ such that $PA=a$ (that is, a circle of radius $a$ centered on $A$) will generally intersect the inscribed circle in two points. These two points have different distances to $B$ and $D$, so the distances $PB$ and $PD$ can't be computed from $A$ (but you can compute a set of two possibilities for each). On the other hand, by symmetry the distance from each of the possible $P$s to $C$ must be the same, so that ought to be calculable.

The exception to this is if $a=\sqrt2\pm1$, in which case there is only one possible location for $P$ (the two circles touch each other rather than intersect).

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