Transformation from circle to ellipse

Statement

Find the transformation (composition of a translation and a dilation) that takes the circle $x^2+y^2=1$ to the ellipse $\dfrac{(x+1)^2}{4}+\dfrac{(y-2)^2}{16}=1$.

Solution

Let us rewrite the ellipse highlighting the squares: $\left(\frac{x+1}{2}\right)^2 + \left(\frac{y-2}{4}\right)^2 = 1.$ We start from the unit circle $u^2 + v^2 = 1$. We apply first the dilation $(x;y)\mapsto(2x;\ 4y)$, which takes the circle to the ellipse centred at the origin $\tfrac{x^2}{4}+\tfrac{y^2}{16}=1$; then the translation by vector $(-1;2)$, that is $(x;y)\mapsto(x-1;\ y+2)$.

Check: composing them, a point $(u;v)$ of the circle goes to $(2u-1;\ 4v+2)$. Setting $X = 2u-1$ and $Y = 4v+2$, we have $u = \tfrac{X+1}{2}$ and $v = \tfrac{Y-2}{4}$, whence $u^2+v^2=1$ gives exactly $\boxed{\frac{(X+1)^2}{4}+\frac{(Y-2)^2}{16}=1.}$

Links

Topics: Plane transformations
Concepts: Circle · Composition of transformations · Dilation · Ellipse · Translation
Methods: Isometric transformations · Homothetic transformations
Skills: Analytic geometry · Using formulae
Exercise type: Geometric problem