Prove that the heat segment joining the 2 centers of the concurrent circles of equal radius is perpendicular to line segment authorized the two intersection point out of the circles. I had come across statements that usual chord that the two circles is bisected if we attract a heat segment involvement the two facility of circles.

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

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

I"ll 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:

$$fracy-k_1x-h_1=frack_2-k_1h_2-h_1$$

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:

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

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.


share
mention
monitor
reply Jul 15 "11 in ~ 3:32
*

J. M. Ain't a civicpride-kusatsu.netematicianJ. M. Ain't a civicpride-kusatsu.netematician
70.6k55 gold badges184184 silver badges333333 bronze badges
$endgroup$
4
include a comment |
1
$egingroup$
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 $AB$.


share
point out
monitor
reply Jul 14 "11 in ~ 21:22
*

Bruno JoyalBruno Joyal
51.5k66 gold badges117117 silver badges216216 bronze title
$endgroup$
include a comment |
1
$egingroup$
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.

See more: 100 Lb Propane Tank Fill Cost, Can You Fill A 100Lb Propane Tank

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.