A circle satisfies Power of a Point, but is the converse true? The "Power of a Point" or "Intersecting Chords" theorem states that for any point in a plane, if a line is drawn that intersects a circle, the distance from the point to one of the intersections multiplied by the distance from the point to the other intersection is a constant for that point, no matter what line you draw, as long as it intersects the circle. My question is if the converse is true. If a shape satisfies this, does it have to be a circle? Are there other shapes that satisfy this?
Here's a neat example: $OA\cdot OB=2$.
Let $P$ be a point and $\Gamma$ be a curve on the plane. A line through $P$ intersects $S$ at two points $A$ and $B$.
I think your statement on 'Power of a Point' can be interpreted in two ways:
1) $PA\cdot PB$ is constant for a fixed $P$. This is true for some $P$ on the plane.
2) $PA\cdot PB$ is constant for a fixed $P$. This is true for every $P$ on the plane.

If you mean (1), then Aretino provides a neat counter-example. In case you want to know the proof:
Let $X = (0,1)$. Note that $OX = OB = 1$.
Then $OA\cdot OB = \frac{1}{\cos{\angle XOA}}\cdot 2(1)(\cos{\angle XOA}) = 2$

If you mean (2), then the converse is indeed true.
Firstly we have to assume for every line through an arbitrary point $P$ that intersects $\Gamma$, it only intersects $\Gamma$ at 2 points (1 if degenerate).
Let another line through $P$ intersect $\Gamma$ at $C$ and $D$. Then $ABCD$ is cyclic. Let its circumcircle be $\Phi$.
Let $F$ be a point on arc $AB$ in $\Phi$ that does not contain $C$. Let $EC$ intersect $\Gamma$ at $E\ne C$, $AB$ at $Q$.
We have $QA\cdot QB = QC\cdot QE$ from given condition, and $QA\cdot QB = QC\cdot QF$ from Power of a Point in $\Phi$.
So, $QE = QF$.
Note that $E$ cannot be on the side of line $AB$ containing $C$, because that will imply $\Gamma$ is concave, and there exists a line that intersects it 3 times.
So, $E = F$. $F$ is a point in $\Gamma$.
Since $F$ can be any point in that arc, the entire arc is also in $\Gamma$.
Choose any $F$ on the arc and repeat the above process. Then we have the entire arc $AB$ that doesn't contain $F$ (i.e. contains $C$) in $\Gamma$.
Therefore, $\Gamma$ consists of the entire circle $\Phi$.
If there exists some point $G$ in $\Gamma$ that isn't in $\Phi$, then draw a line through $G$ and the centre of $\Phi$. Then the line intersects $\Gamma$ 3 times (once at $G$, twice at $\Phi$).
Thus, $\Gamma$ = $\Phi$ is a circle.

(Sorry for no diagram; couldn't draw it right.)