I need a simple proof that a line cannot intersect a circle at three distinct points.
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$\begingroup$ With algebra, you can substitute in the equation of the circle, get a quadratic. But then you have to show a quadratic cannot have more than $2$ roots. $\endgroup$– André NicolasMar 1, 2012 at 17:50
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6$\begingroup$ Counterexample: the unit circle intersects the $y$-axis in $4$ points $(0,\pm1),\:(0,\pm4)$ over $\mathbb Z/15\qquad$ $\endgroup$– Bill DubuqueMar 1, 2012 at 22:36
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$\begingroup$ @BillDubuque i did't understand how unit circle intersects the x axis in 4points. $\endgroup$– SaurabhJun 4, 2012 at 20:02
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$\begingroup$ @SaurabhHota $\rm\:x^2 + y^2 = 1\:$ has solutions $\rm\:(x,y) = (0,\pm1),\ (0,\pm4)\:$ in integers modulo $15.$ $\endgroup$– Bill DubuqueJun 4, 2012 at 20:30
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4 Answers
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Or a more geometric proof: If a circle intersects a line in $A$ and $B$, the center of the circle lies on the center normal of the line segment $AB$. If there is a third intersection point $C$, the center of the circle must also lie on the center normal of $BC$. But these two center normals are distinct parallel lines, and cannot have point in common.
answered Mar 1, 2012 at 17:54
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Without loss of generality, assume the circle is $x^2 + y^2 = r^2$ and the line is $y = mx + c$.
The x coordinates of the point of intersection satisfy $x^2 + (mx+c)^2 = r^2$ which is a quadratic and hence has at most $2$ roots.
Since given an $x$, the $y$ on the line is uniquely determined, we are done.
J. M. ain't a mathematician
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answered Mar 1, 2012 at 17:51
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1$\begingroup$ I always read “with loss of generality” when I see “wlog”. So I use “wolog” instead. But in some sense, “wlog” is more correct. $\endgroup$ Mar 1, 2012 at 17:56
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$\begingroup$ @HaraldHanche-Olsen: You are right, there is ambiguity there :-) $\endgroup$ Mar 1, 2012 at 17:58
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Similar to Harald's proof, draw in a radius from the center of the circle to each point where the line intersects the circle. Now draw a perpendicular segment from the center to a point C on the line. Assuming we have more than one point of intersection, we have multiple right triangles which are congruent due to the HL theorem. Clearly we can't have a third point of intersection because there cannot be 3 distinct points along the line equidistant from C.
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Take any 3 distinct points on a circle and notice that each angle of the triangle formed by those 3 points is higher than 0 and smaller than 180 degrees. Any of the angles formed by 3 distinct points on a line (degenerate triangle) takes a value of either 0 degrees or 180 degrees.
The proof is complete.
answered Jun 4, 2012 at 19:48
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$\begingroup$ It seesm to me that you postulate what you want to prove. why are all angles of suche a triangle greater than 0 and smaller then 180 degrees? $\endgroup$ Sep 20, 2013 at 19:39
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