# Prove that every rigid motion transforms circles into circles

• Feb 11th 2010, 05:35 PM
Pinkk
Prove that every rigid motion transforms circles into circles
• Feb 11th 2010, 05:39 PM
Drexel28
Quote:

Originally Posted by Pinkk

What's your definition of rigid motion?
• Feb 11th 2010, 05:46 PM
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?
• Feb 11th 2010, 05:54 PM
Drexel28
Quote:

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 :p. 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 $x\in f(C)$.
• 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 $O$. 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!