Prove that every rigid motion transforms circles into circles

**Pinkk** · February 11th 2010, 05:35 PM

I do not know how to go about this. Any help would be appreciated.

**Drexel28** · February 11th 2010, 05:39 PM

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?

**Pinkk** · February 11th 2010, 05:46 PM

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

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

**Drexel28** · February 11th 2010, 05:54 PM

Originally Posted by Pinkk

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

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

Oh...and isometry

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

**Pinkk** · February 11th 2010, 06:00 PM

Originally Posted by Drexel28

Oh...and isometry

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

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!

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