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Prove that the line segment joining the two centers of the concurrent circles of equal radius is perpendicular to line segment joining the two intersection points of the circles. I had come across statements that common chord of the two circles is bisected if we draw a line segment joining the two center of circles.

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I was told that. I should use some property of rhombus to prove above statment.what property is that? – user13359 Jul 15 '11 at 18:31
I have assumed that since two circles are of equal radius and intersect each other they are "concurrent circles". Is this term correct or is their any other term for this type of circles? – user13359 Jul 15 '11 at 18:34
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The argument by symmetry given by Joriki is in my opinion optimal.

If we want a maximally "high school" proof of the old-fashioned type, let the centers of the circles be $C$ and $C'$, and let the intersection points of the circles be $A$ and $B$, as in the answer by Bruno Joyal. Let $M$ be the point where lines $AB$ and $CC'$ meet.

Then triangles $ACC'$ and $BCC'$ are congruent, by what used to be called SSS.

It follows that $\angle ACC'=\angle BCC'$, so triangles $ACM$ and $BCM$ are congruent, by what used to be called SAS.

Thus $\angle CMA=\angle CMB$. But these angles add up to a "straight angle," so each is a right angle.

Comment: We have used the past tense repeatedly in the proof, since in many school curricula geometry of the classical kind has died. In the death spiral, students were made to answer multiple choice questions about "two-column" proofs.

Added: Curiously enough, we all seem to have missed the "equal radius" part of the statement of the problem, which as far as I can tell was there from the beginning! Of course, the result is true without the assumption of equal radius.

But the following argument may have been the intended one. Since the radii are equal, the centers of the circles and their points of intersection form a rhombus. By a theorem which has presumably been proved in the course prior to this problem, the diagonals of a rhombus are perpendicular (and bisect each other). End of proof.

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This is true by symmetry. The figure is symmetric with respect to reflection in the line joing the centres; if the line joining the two intersection points weren't perpendicular to that line, it would break the symmetry.

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Let the points of intersection of the circles be $A$ and $B$, and the centers be $C$ and $C'$. The triangles $ACB$ and $AC'B$ are two isoceles triangles which share the same base. As you have said, the line joining $C$ and $C'$ bisects this base. But in an isoceles triangle, the main altitude is the same line as the main median. Hence the line $CC'$ must be an altitude for each of the triangles $ACB$ and $AC'B$, i.e., it is perpendicular to $AB$.

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Here's a coordinate geometry proof: let the two circles be of the form

$$(x-h_1)^2+(y-k_1)^2=r_1^2,\qquad (x-h_2)^2+(y-k_2)^2=r_2^2$$

I'll be demonstrating a more general statement: the radical line of the two circles is perpendicular to the line through the two centers. In the special case of intersecting circles, the radical line is the line through the two intersection points.

It is trivial to write down the equation of the line joining the two centers:


To construct the equation of the radical line, we expand the Cartesian equations of the two 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 from the other:


whose slope is $-\frac{h_2-h_1}{k_2-k_1}$, which when multiplied by the slope $\frac{k_2-k_1}{h_2-h_1}$ of the line joining the centers gives $-1$, thus showing the perpendicularity.

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Good demonstration of the "algorithmic" nature of coordinate geometry arguments. – André Nicolas Jul 15 '11 at 4:34
Indeed, @André (meta: it is nice to be calling you that instead of "6312" :) ). On the other hand, there remains a sort of panache when one goes the classical route instead of the coordinate route... – J. M. Jul 16 '11 at 4:51
Euclid is said to have told a Ptolemy, who was asking for shortcuts to proofs, and/or Menaechmus told Alexander "There is no royal road to geometry." But there is, for a king with a large enough computing budget. And the anecdote is unlikely to be true. It is unhealthy to dis a king; one may lose one's grant, or worse. – André Nicolas Jul 16 '11 at 6:01
Yes @André, I could have made the centers lie on the horizontal axis, and the proof certainly becomes much tidier that way... maybe I'll rewrite later. – J. M. Jul 16 '11 at 6:49

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