How to convert points in a circle to points in an ellipse I have a circle with points(X, Y) inside it that I want to be able to adjust to equivalent points in an ellipse with equivalent width. How can I do this?
I understand that an ellipse with equal height and width is a circle and that the only thing that I change is the height but as a result of this, there is surely also some change in width at points that need to be taken into account for converting my points.
 A: The points $(x, ky)$ for any real number $k$ will give you an ellipse.
A circle is defined by the points satisfying $x^2+y^2=r^2$. Replacing $y$ with $ky$ yields the formula $x^2+k^2y^2=r^2$. Dividing both sides by $r^2$ yields $\frac{x^2}{r^2}+\frac{k^2y^2}{r^2} =1$, an ellipse in standard form.
A: The parametric equation of a circle with center $ (x_0, y_0)$ and radius $r$ is $$ x=x_0 +r cos(\theta ) , y=y_0 +r sin(\theta ) $$
The parametric equation of an ellipse with center $ (x_0, y_0)$ and semi axes lengths $a$ and $b$ is $$ x=x_0 +a cos(\theta ) , y=y_0 +b sin(\theta ) $$
Thus transformation from a circle to an ellipse is scaling $ x-x_0$ and   $ y-y_0.$
A: HINT.-Any circle is in bijective correspondence (actually is homeomorphic) with all ellipse having the same center (and a lot of others ellipses). See for example the circle $x^2+y^2=4$ and the ellipse$\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$. Then, for example, the ligne $y=\dfrac x2$ defines some of these bijections (the analytic expressions are easy to calculate).

