You are watching: Segment joining two points on a circle
This is true by symmetry. The figure is symmetric through respect come reflection in the heat joing the centres; if the heat joining the two intersection points weren"t perpendicular to that line, it would certainly break the symmetry.
Here"s a name: coordinates geometry proof: allow the two circles be of the form
I"ll it is in demonstrating a much more general statement: the radical line of the 2 circles is perpendicular to the line through the two centers. In the special case of intersecting circles, the radical line is the line with the two intersection points.
It is trivial to compose down the equation that the heat joining the 2 centers:
To construct the equation the the radical line, we increase the Cartesian equations the the 2 circles:
$$x^2-2h_1 x+h_1^2+y^2-2k_1 y+k_1^2=r_1^2,qquad x^2-2h_2 x+h_2^2+y^2-2k_2 y+k_2^2=r_2^2$$
and then subtract one native the other:
whose slope is $-frach_2-h_1k_2-k_1$, which as soon as multiplied by the slope $frack_2-k_1h_2-h_1$ of the line joining the centers provides $-1$, for this reason showing the perpendicularity.
reply Jul 15 "11 in ~ 3:32
J. M. Ain't a civicpride-kusatsu.netematicianJ. M. Ain't a civicpride-kusatsu.netematician
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Let the points of intersection the the one be $A$ and also $B$, and also the centers be $C$ and also $C'$. The triangles $ACB$ and $AC'B$ room two isoceles triangles which re-publishing the same base. Together you have said, the line joining $C$ and $C'$ bisects this base. But in one isoceles triangle, the main altitude is the same line as the main median. Therefore the line $CC'$ should be an altitude for each of the triangles $ACB$ and also $AC'B$, i.e., it is perpendicular come $AB$.
reply Jul 14 "11 in ~ 21:22
Bruno JoyalBruno Joyal
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The debate by symmetry given by Joriki is in my opinion optimal.
If we want a maximally "high school" proof of the old-fashioned type, permit the centers that the one be $C$ and also $C'$, and let the intersection points of the circles be $A$ and $B$, together in the answer by Bruno Joyal. Let $M$ be the suggest where currently $AB$ and also $CC'$ meet.
Then triangles $ACC'$ and $BCC'$ room congruent, by what used to be dubbed SSS.
It complies with that $angle ACC'=angle BCC'$, so triangle $ACM$ and also $BCM$ space congruent, by what used to be dubbed SAS.
Thus $angle CMA=angle CMB$. However these angles add up to a "straight angle," so each is a appropriate angle.
Comment: We have actually used the past tense repeatedly in the proof, due to the fact that in countless school curricula geometry the the classic kind has actually died. In the death spiral, students were made come answer multiple selection questions about "two-column" proofs.
Added: Curiously enough, us all it seems to be ~ to have actually missed the "equal radius" part of the declare of the problem, i m sorry as much as I have the right to tell to be there from the beginning! of course, the an outcome is true without the assumption of same radius.
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But the following debate may have been the to plan one. Since the radii are equal, the centers the the circles and also their point out of intersection kind a rhombus. By a theorem which has actually presumably been showed in the course before this problem, the diagonals the a rhombus space perpendicular (and bisect each other). End of proof.