# Prove that every rigid motion transforms circles into circles

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• Feb 11th 2010, 05:35 PM
Pinkk
Prove that every rigid motion transforms circles into circles
I do not know how to go about this. Any help would be appreciated.
• Feb 11th 2010, 05:39 PM
Drexel28
Quote:

Originally Posted by Pinkk
I do not know how to go about this. Any help would be appreciated.

What's your definition of rigid motion?
• Feb 11th 2010, 05:46 PM
Pinkk
A rigid motion of the Euclidean plane is a transformation $\displaystyle f(P)$ of the plane into itself such that $\displaystyle d(f(P),f(Q))=d(P,Q)$.

Would I just write that if you are given a circle $\displaystyle O$ with center $\displaystyle A$ and any arbitrary point on the circle $\displaystyle B$ that is always distance $\displaystyle \alpha$ from the center, then according to the definition of rigid motion, $\displaystyle d(f(A),f(B))=d(A,B)=\alpha$, so any arbitrary point $\displaystyle B'$ on $\displaystyle O'$ is distance $\displaystyle \alpha$ from $\displaystyle A'$, and so $\displaystyle O'$ is a circle?
• Feb 11th 2010, 05:54 PM
Drexel28
Quote:

Originally Posted by Pinkk
A rigid motion of the Euclidean plane is a transformation $\displaystyle f(P)$ of the plane into itself such that $\displaystyle d(f(P),f(Q))=d(P,Q)$.

Would I just write that if you are given a circle $\displaystyle O$ with center $\displaystyle A$ and any arbitrary point on the circle $\displaystyle B$ that is always distance $\displaystyle \alpha$ from the center, then according to the definition of rigid motion, $\displaystyle d(f(A),f(B))=d(A,B)=\alpha$, so any arbitrary point $\displaystyle B'$ on $\displaystyle O'$ is distance $\displaystyle \alpha$ from $\displaystyle A'$, and so $\displaystyle O'$ is a circle?

Oh...and isometry :p. A set of points forms a circle if all the points are equidistant from a signle point $\displaystyle O$. So we know that $\displaystyle d(O,x)=r$ for every $\displaystyle x\in C$ (our circle) and so $\displaystyle d\left(f(x),f(O)\right)=d(x,O)=r$ for every $\displaystyle x\in f(C)$.
• Feb 11th 2010, 06:00 PM
Pinkk
Quote:

Originally Posted by Drexel28
Oh...and isometry :p. A set of points forms a circle if all the points are equidistant from a signle point $\displaystyle O$. So we know that $\displaystyle d(O,x)=r$ for every $\displaystyle x\in C$ (our circle) and so $\displaystyle d\left(f(x),f(O)\right)=d(x,O)=r$ for every $\displaystyle x\in f(C)$.

Ah okay. I had a vague idea that that must have been the approach, but I was having a problem formulating it into an actual proof. Thanks!