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This is not a complete answer, but just some extended comments:

1. Existence of an inflection point inside every circle

Suppose that $\gamma$ is $C^2$ so that its curvature is everywhere well-defined. Then $\gamma$ must have an inflection point, i.e., a point of vanishing curvature, inside every one of these circles. If not, then the segment $S$ of $\gamma$ contained in one of the circles, say $C$, is locally convex. Since the end points of $S$ lie on $C$, it follows that $S$ is convex. Indeed, by gluing to $S$ one of the segments of $C$ determined by the end points of $S$, we obtain a simple closed locally convex curve, and every such curve must be globally convex, i.e., bound a convex set. So $S$ must lie strictly on one side of the tangent line of $\gamma$ passing through the center of $C$. Hence $S$ cannot bisect the area of the region bounded by $C$.

The above argument shows, in particular, that if the area bisecting property holds for circles of arbitrary small radius, then the curvature of $\gamma$ must vanish identically, which means that $\gamma$ has to be a line.

2. A generelized conjecture

It seems that a more general phenomenon could be true. Suppose that $\gamma$ divides the circles (of some fixed radius centered at points of $\gamma$) into regions of constant area. Then one might conjecture that $\gamma$ should be a circle. The radius of the circle is determined by the ratio of the areas of the two regions. When the two regions have equal area, then the radius is infinite or $\gamma$ is a line.

3. A local symmetry condition

If $\gamma$ is $C^1$, then one might try to prove the generalized conjecture proposed in item 2 above via a local symmetry condition as follows. Suppose that $\gamma$ passes through the origin $o$ of $R^2$ and is tangent to the $x$-axis at $o$. Further suppose that the circles have radii $1$. So $\gamma$ bisects the area of $S^1$. Let $A$ be the area of one of the regions, say $\Omega$, determined by $\gamma$ inside $S^1$. The rate of change of $A$, as the circle moves along $\gamma$, is determined by the segment $\Omega\cap S^1$. More specifically, the rate of change of $A$ depends on the portion of $\Omega\cap S^1$ in the right semicircle of $S^1$ versus the portion of $\Omega\cap S^1$ in the left semicircle of $S^1$. These portions must have equal length, since the rate of change of $A$ is zero. So if we let $p_+$ and $p_-$ be the intersection points of $\gamma$ with $S^1$, then $p_+$ and $p_-$ will be symmetric with respect to the $y$-axis.

More generally, for any point $p$ of $\gamma$ let $p_+$ and $p_-$ be the intersection of $\gamma$ with the circle centered at $p$. Then the normal line of $\gamma$ at $p$ must be orthogonal to and bisect the segment $p_+p_-$. I think that this condition might characterize circles, and it will force $\gamma$ to be a line when $p$, $p_+$, and $p_-$ are collinear.

This is not a complete answer, but just some extended comments:

1. Existence of an inflection point inside every circle

Suppose that $\gamma$ is $C^2$ so that its curvature is everywhere well-defined. Then $\gamma$ must have an inflection point, i.e., a point of vanishing curvature, inside every one of these circles. If not, then the segment $S$ of $\gamma$ contained in one of the circles, say $C$, is locally convex. Since the end points of $S$ lie on $C$, it follows that $S$ is convex. Indeed, by gluing to $S$ one of the segments of $C$ determined by the end points of $S$, we obtain a simple closed locally convex curve. Every such curve must be globally convex, i.e., bound a convex set. So $S$ must lie strictly on one side of the tangent line of $\gamma$ passing through the center of $C$. Hence $S$ cannot bisect the area of the region bounded by $C$.

The above argument shows, in particular, that if the area bisecting property holds for circles of arbitrary small radius, then the curvature of $\gamma$ must vanish identically, which means that $\gamma$ has to be a line.

2. A generelized conjecture

It seems that a more general phenomenon could be true. Suppose that $\gamma$ divides the circles (of some fixed radius centered at points of $\gamma$) into regions of constant area. Then one might conjecture that $\gamma$ should be a circle. The radius of the circle is determined by the ratio of the areas of the two regions. When the two regions have equal area, then the radius is infinite or $\gamma$ is a line.

3. A local symmetry condition

If $\gamma$ is $C^1$, then one might try to prove the generalized conjecture proposed in item 2 above via a local symmetry condition as follows. Suppose that $\gamma$ passes through the origin $o$ of $R^2$ and is tangent to the $x$-axis at $o$. Further suppose that the circles have radii $1$. So $\gamma$ bisects the area of $S^1$. Let $A$ be the area of one of the regions, say $\Omega$, determined by $\gamma$ inside $S^1$. The rate of change of $A$, as the circle moves along $\gamma$, is determined by the segment $\Omega\cap S^1$. More specifically, the rate of change of $A$ depends on the portion of $\Omega\cap S^1$ in the right semicircle of $S^1$ versus the portion of $\Omega\cap S^1$ in the left semicircle of $S^1$. These portions must have equal length, since the rate of change of $A$ is zero. So if we let $p_+$ and $p_-$ be the intersection points of $\gamma$ with $S^1$, then $p_+$ and $p_-$ will be symmetric with respect to the $y$-axis.

More generally, for any point $p$ of $\gamma$ let $p_+$ and $p_-$ be the intersection of $\gamma$ with the circle centered at $p$. Then the normal line of $\gamma$ at $p$ must be orthogonal to and bisect the segment $p_+p_-$. I think that this condition might characterize circles, and it will force $\gamma$ to be a line when $p$, $p_+$, and $p_-$ are collinear.

This is not a complete answer, but just some extended comments:

Suppose that $\gamma$ is $C^2$ so that its curvature is everywhere well-defined. Then $\gamma$ must have an inflection point, i.e., a point of vanishing curvature, inside every one of these circles. If not, then the segment $S$ of $\gamma$ contained in one of the circles, say $C$, is locally convex. Since the end points of $S$ lie on $C$, it follows that $S$ is convex. Indeed, by gluing to $S$ one of the segments of $C$ determined by the end points of $S$, we obtain a simple closed locally convex curve, and every such curve must be globally convex, i.e., bound a convex set. So $S$ must lie strictly on one side of the tangent line of $\gamma$ passing through the center of $C$. Hence $S$ cannot bisect the area of the region bounded by $C$.

The above argument shows, in particular, that if the area bisecting property holds for circles of arbitrary small radius, then the curvature of $\gamma$ must vanish identically, which means that $\gamma$ has to be a line.

It seems that a more general phenomenon could be true. Suppose that $\gamma$ divides the circles (of some fixed radius centered at points of $\gamma$) into regions of constant area. Then one might conjecture that $\gamma$ should be a circle. The radius of the circle is determined by the ratio of the areas of the two regions. When the two regions have equal area, then the radius is infinite or $\gamma$ is a line.

If $\gamma$ is $C^1$, then one might try to prove the generalized conjecture proposed in item 2 above via a local symmetry condition as follows. Suppose that $\gamma$ passes through the origin $o$ of $R^2$ and is tangent to the $x$-axis at $o$. Further suppose that the circles have radii $1$. So $\gamma$ bisects the area of $S^1$. Let $A$ be the area of one of the regions, say $\Omega$, determined by $\gamma$ inside $S^1$. The rate of change of $A$, as the circle moves along $\gamma$, is determined by the segment $\Omega\cap S^1$. More specifically, the rate of change of $A$ depends on the portion of $\Omega\cap S^1$ in the right semicircle of $S^1$ versus the portion of $\Omega\cap S^1$ in the left semicircle of $S^1$. These portions must have equal length, since the rate of change of $A$ is zero. So if we let $p_+$ and $p_-$ be the intersection points of $\gamma$ with $S^1$, then $p_+$ and $p_-$ will be symmetric with respect to the $y$-axis.

More generally, for any point $p$ of $\gamma$ let $p_+$ and $p_-$ be the intersection of $\gamma$ with the circle centered at $p$. Then the normal line of $\gamma$ at $p$ must be orthogonal to and bisect the segment $p_+p_-$. I think that this condition might characterize circles, and it will force $\gamma$ to be a line when $p$, $p_+$, and $p_-$ are collinear.

This is not a complete answer, but just some extended comments:

1. Existence of an inflection point inside every circle

Suppose that $\gamma$ is $C^2$ so that its curvature is everywhere well-defined. Then $\gamma$ must have an inflection point, i.e., a point of vanishing curvature, inside every one of these circles. If not, then the segment $S$ of $\gamma$ contained in one of the circles, say $C$, is locally convex. Since the end points of $S$ lie on $C$, it follows that $S$ is convex. Indeed, by gluing to $S$ one of the segments of $C$ determined by the end points of $S$, we obtain a simple closed locally convex curve, and every such curve must be globally convex, i.e., bound a convex set. So $S$ must lie strictly on one side of the tangent line of $\gamma$ passing through the center of $C$. Hence $S$ cannot bisect the area of the region bounded by $C$.

The above argument shows, in particular, that if the area bisecting property holds for circles of arbitrary small radius, then the curvature of $\gamma$ must vanish identically, which means that $\gamma$ has to be a line.

2. A generelized conjecture

It seems that a more general phenomenon could be true. Suppose that $\gamma$ divides the circles (of some fixed radius centered at points of $\gamma$) into regions of constant area. Then one might conjecture that $\gamma$ should be a circle. The radius of the circle is determined by the ratio of the areas of the two regions. When the two regions have equal area, then the radius is infinite or $\gamma$ is a line.

3. A local symmetry condition

If $\gamma$ is $C^1$, then one might try to prove the generalized conjecture proposed in item 2 above via a local symmetry condition as follows. Suppose that $\gamma$ passes through the origin $o$ of $R^2$ and is tangent to the $x$-axis at $o$. Further suppose that the circles have radii $1$. So $\gamma$ bisects the area of $S^1$. Let $A$ be the area of one of the regions, say $\Omega$, determined by $\gamma$ inside $S^1$. The rate of change of $A$, as the circle moves along $\gamma$, is determined by the segment $\Omega\cap S^1$. More specifically, the rate of change of $A$ depends on the portion of $\Omega\cap S^1$ in the right semicircle of $S^1$ versus the portion of $\Omega\cap S^1$ in the left semicircle of $S^1$. These portions must have equal length, since the rate of change of $A$ is zero. So if we let $p_+$ and $p_-$ be the intersection points of $\gamma$ with $S^1$, then $p_+$ and $p_-$ will be symmetric with respect to the $y$-axis.

More generally, for any point $p$ of $\gamma$ let $p_+$ and $p_-$ be the intersection of $\gamma$ with the circle centered at $p$. Then the normal line of $\gamma$ at $p$ must be orthogonal to and bisect the segment $p_+p_-$. I think that this condition might characterize circles, and it will force $\gamma$ to be a line when $p$, $p_+$, and $p_-$ are collinear.

This is not a complete answer, but just some extended comments:

Suppose that $\gamma$ is $C^2$ so that its curvature is everywhere well-defined. Then $\gamma$ must have an inflection point, i.e., a point of vanishing curvature, inside every one of these circles. If not, then the segment $S$ of $\gamma$ contained in one of the circles, say $C$, is locally convex. Since the end points of $S$ lie on $C$, it follows that $S$ is convex. Indeed, by gluing to $S$ one of the segments of $C$ determined by the end points of $S$, we obtain a simple closed locally convex curve, and every such curve must be globally convex, i.e., bound a convex set. So $S$ must lie strictly on one side of the tangent line of $\gamma$ passing through the center of $C$. Hence $S$ cannot bisect the area of the region bounded by $C$.

The above argument shows, in particular, that if the area bisecting property holds for circles of arbitrary small radius, then the curvature of $\gamma$ must vanish identically, which means that $\gamma$ has to be a line.

It seems that a more general phenomenon could be true. Suppose that $\gamma$ divides the circles (of some fixed radius centered at points of $\gamma$) into regions of constant area. Then one might conjecture that $\gamma$ should be a circle. The radius of the circle is determined by the ratio of the areas of the two regions. When the two regions have equal area, then the radius is infinite or $\gamma$ is a line.

If $\gamma$ is $C^1$, then one might try to prove the generalized conjecture proposed in item 2 above via a local symmetry condition as follows. Suppose that $\gamma$ passes through the origin $o$ of $R^2$ and is tangent to the $x$-axis at $o$. Further suppose that the circles have radii $1$. So $\gamma$ bisects the area of $S^1$. Let $A$ be the area of one of the regions, say $\Omega$, determined by $\gamma$ inside $S^1$. The rate of change of $A$, as the circle moves along $\gamma$, is determined by the segment $\Omega\cap S^1$. More specifically, the rate of change of $A$ depends on the portion of $\Omega\cap S^1$ in the right semicircle of $S^1$, and the portion of $\Omega\cap S^1$ in the left semicircle of $S^1$. These portions must have equal length, since the rate of change of $A$ is zero. So if we let $p_+$ and $p_-$ be the intersection points of $\gamma$ with $S^1$, then $p_+$ and $p_-$ will be symmetric with respect to the $y$-axis.

More generally, for any point $p$ of $\gamma$ let $p_+$ and $p_-$ be the intersection of $\gamma$ with the circle centered at $p$. Then the normal line of $\gamma$ at $p$ must be orthogonal to and bisect the segment $p_+p_-$. I think that this condition might characterize circles, and it will force $\gamma$ to be a line when $p$, $p_+$, and $p_-$ are collinear.

This is not a complete answer, but just some extended comments:

Suppose that $\gamma$ is $C^2$ so that its curvature is everywhere well-defined. Then $\gamma$ must have an inflection point, i.e., a point of vanishing curvature, inside every one of these circles. If not, then the segment $S$ of $\gamma$ contained in one of the circles, say $C$, is locally convex. Since the end points of $S$ lie on $C$, it follows that $S$ is convex. Indeed, by gluing to $S$ one of the segments of $C$ determined by the end points of $S$, we obtain a simple closed locally convex curve, and every such curve must be globally convex, i.e., bound a convex set. So $S$ must lie strictly on one side of the tangent line of $\gamma$ passing through the center of $C$. Hence $S$ cannot bisect the area of the region bounded by $C$.

The above argument shows, in particular, that if the area bisecting property holds for circles of arbitrary small radius, then the curvature of $\gamma$ must vanish identically, which means that $\gamma$ has to be a line.

It seems that a more general phenomenon could be true. Suppose that $\gamma$ divides the circles (of some fixed radius centered at points of $\gamma$) into regions of constant area. Then one might conjecture that $\gamma$ should be a circle. The radius of the circle is determined by the ratio of the areas of the two regions. When the two regions have equal area, then the radius is infinite or $\gamma$ is a line.

If $\gamma$ is $C^1$, then one might try to prove the generalized conjecture proposed in item 2 above via a local symmetry condition as follows. Suppose that $\gamma$ passes through the origin $o$ of $R^2$ and is tangent to the $x$-axis at $o$. Further suppose that the circles have radii $1$. So $\gamma$ bisects the area of $S^1$. Let $A$ be the area of one of the regions, say $\Omega$, determined by $\gamma$ inside $S^1$. The rate of change of $A$, as the circle moves along $\gamma$, is determined by the segment $\Omega\cap S^1$. More specifically, the rate of change of $A$ depends on the portion of $\Omega\cap S^1$ in the right semicircle of $S^1$ versus the portion of $\Omega\cap S^1$ in the left semicircle of $S^1$. These portions must have equal length, since the rate of change of $A$ is zero. So if we let $p_+$ and $p_-$ be the intersection points of $\gamma$ with $S^1$, then $p_+$ and $p_-$ will be symmetric with respect to the $y$-axis.

More generally, for any point $p$ of $\gamma$ let $p_+$ and $p_-$ be the intersection of $\gamma$ with the circle centered at $p$. Then the normal line of $\gamma$ at $p$ must be orthogonal to and bisect the segment $p_+p_-$. I think that this condition might characterize circles, and it will force $\gamma$ to be a line when $p$, $p_+$, and $p_-$ are collinear.