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CTK Exchange
confused
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Jun-04-06, 09:54 AM (EST) |
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"Proofs using two circles"
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Two circles with centers X and Y intersect at points P and Q. Prove that XY is the perpendicular bisector of PQ. My assignment is to prove the above statement in as many different ways as possible... using analytic methods, vector methods, converse statments, congruent triangles, similar triangles... anything that works. The circles in the proof should be two different sizes, but if I am really stuck, I can make them the same size.
I have tried vector proofs using the dot product. I could not find the proper way to start because there are so many ways to expand the dot product. I only ended up proving my opening statement that I was using to prove that XY was the perpendicular bisector of PQ.
I have also tried analytic methods, but I could not easily find the intersection points of the two circles.
Please help. I am not strong in proofs when it comes to circles. I do not know enough properties of circles to be able to use vector methods and I am confused as to where to start this problem. My teacher wants me to find the hardest methods to prove it as possible.
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alexb
Charter Member
1971 posts |
Jun-04-06, 09:59 AM (EST) |
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1. "RE: Proofs using two circles"
In response to message #0
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>Two circles with centers X and Y intersect at points P and
>Q. Prove that XY is the perpendicular bisector of PQ.
>
>My assignment is to prove the above statement in as many
>different ways as possible... using analytic methods, vector
>methods, converse statments, congruent triangles, similar
>triangles... anything that works. The circles in the proof
>should be two different sizes, but if I am really stuck, I
>can make them the same size. Two advices:
- Draw a picture and have a good look at it.
- Do not worry about doing that in many different ways. Do it just once. You'll be able to fix this later.
>
>My teacher wants me to find the hardest
>methods to prove it as possible. Since you are stuck, any proof will be as difficult as possible. |
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jay_shark
Member since Mar-20-07
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Mar-20-07, 01:48 PM (EST) |
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2. "RE: Proofs using two circles"
In response to message #1
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Proof 1: Notice that triangle pxy is congruent to qxy .
Since xy is common .
xp=xq and py=qy so these triangles are congruent using SSS . Call the point where pq intersects xy =w .
Now it's easy to show pxw is congruent to qxw using SAS since angle pxy = angle qxy . We know that the angle pwq=180 since it's a straight line.
pwx=qxw and pxw+qxy=180 which implies that pxw=qxw=90 . Proof 2: Using a co-ordinate system . WLOG suppose circle one is centered at the origin with equation x^2+y^2=r^2 Circle two has equation (x-a)^2 +y^2 =r'^2 we have two equations with constants r,r' and a eq 2 - 1 ; (x-a)^2 -x^2 = r'^2 -r^2
x^2 -2xa +a^2 -x^2 =r'^2-r^2
-2xa + a^2 = r'^2 -r^2
a^2-r'^2-r^2= 2xa
(a^2-r'^2-r^2)/2a=x plug this value of x to either equation 1 or 2 and solve for y . You will get two different values for y (+ -) . Now verify that x^2+y^2=r^2 from the pythagorean theorem |
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