Question

Take a circle and a line. Shear the entire picture. Does the shortest distance between the circle and the line get mapped by the shearing transformation to the shortest distance between the ellipse and line?

Answer 1

How do you transform the shortest distance, which is a scalar, under a linear transformation, which only applies to vectors?

Answer 2

I was really just too lazy to write out "path of shortest distance"...

Answer 3

I am really tempted to say no. Consider a circle centered in the origin. Any vector from the origin to the circle is a shortest path but after shearing most of them are not thr shortest anymore. I know origin is not a line but it is really hard to conceive that it doesn't work for a point but it does for a line.

Answer 4

Consider the circle \(x^2+y^2=1\) and line \(\lambda\begin{pmatrix}1\\-1\end{pmatrix}+\begin{pmatrix}1.5\\1.5\end{pmatrix}\). The matrix representation of the circle is\[
\mathbf{x}^TA\mathbf{x}=0\\
A=
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{pmatrix}
\]in homogeneous coordinates. Under the shear transformation \(M=\begin{pmatrix}1&2\\0&1\end{pmatrix}\), the coefficient matrix \(A\) transform as \(M'^TAM'=\begin{pmatrix}1&2&0\\2&5&0\\0&0&-1\end{pmatrix}\), corresponding to \(x^2+4xy+5y^2-1=0\), \(M'=\begin{pmatrix}1&2&0\\0&1&0\\0&0&1\end{pmatrix}\). The line transform to \(\lambda\begin{pmatrix}-1\\-1\end{pmatrix}+\begin{pmatrix}4.5\\1.5\end{pmatrix}\). The transformed line and the ellipse intersects despite the circle and the original line did not. Furthermore, the closest point on the original line to the circle is \(\begin{pmatrix}1.5\\1.5\end{pmatrix}\) by construction but it transforms to \(\begin{pmatrix}4.5\\1.5\end{pmatrix}\), which is not one of the two points in which the transformed line intersects the ellipse.

Answer 5

Nope, this is incorrect. The shear transformation from circle to ellipse in homogeneous coordinate is not this.

Answer 6

Somehow the matrix \(\begin{pmatrix}1&2&0\\0&1&0\\0&0&1\end{pmatrix}\) reflected the ellipse upside down.

Answer 7

Apparently the coefficient matrix transform by \(A\mapsto \left(M^{-1}\right)^TAM^{-1}\). The graph from the correct transformation looks like your assertion is true...