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I've just understood that any two circles can't have more then 2 points of intersection, drawing several different intersected circles.

Is there a way to prove it only by "geometric constructions" method, using only a compass and a ruler without divisions ?

curioushuman
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    Refer to https://math.stackexchange.com/questions/16634/prove-that-three-points-are-enough-to-draw-define-one-and-only-one-circle – auntyellow Jul 08 '22 at 12:36
  • Select 2 points of intersection $A, B$, Draw the line which perpendicular to the line segment $AB$ and coming from it's half. Repeat with another 2 points of intersection . Then the 2 lines Cross at a unique point, the center of the circles. Also the 2 circles sjoukd have the same radius. This is enough for uniqueness – dmtri Jul 08 '22 at 12:37
  • Let center of first circle is $O_1$, radius of first circle is $R_1$, center of second circle is $O_2$, radius of second circle if $R_2$, distance $O_1O_2=d$. All intersection points $X_i$ should satisfy $X_iC_1=R_1$, $X_iC_2=R_2$. Then for all points $X_i$ triangles $X_iC_1C_2$ are congruent by 3 sides. Let draw any triangle $ABC$ with sides $AB=d$, $AC=R_1$, $BC=R_2$. Then angle $X_iC_1C_2$ must be congruent to angle $CAB$. Using compass and ruler we can draw only two rays $C_1X_i$ such that $X_iC_1C_2=CAB$. For both two rays there is only one $X_i$ such that $X_iC_1=R_1$. – Ivan Kaznacheyeu Jul 11 '22 at 12:45
  • Briefly, construction itself only cannot prove some claim, one needs additional facts. In my previous comment I've used equivalent fact that there is only one ray having given angle with given ray counted in given direction. – Ivan Kaznacheyeu Jul 11 '22 at 12:47

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