Find the image of the circle under the Transformation?

Question

Find the image of the circle $$x^2 + y^2 = 4$$ under the transformation $$T ((x,y)) = (x,y + 2x)$$

Answer 1

Let $$T ((x,y))=(X,Y)$$$$\therefore X=x$$$$Y=y+2x$$$$\Rightarrow Y=y+2X$$ $$\Rightarrow y=Y-2X$$Plugging in the original equation $$X^2+(Y-2X)^2=4$$$$\Rightarrow 5X^2-4XY+Y^2=4$$

Answer 2

A linear transform maps the unit circle on an ellipse.
As a start: To make your life easier, first transform the unit circle.
Since it is a linear transform, the image of the circle with radius 2 will be twice the size of the image of the one with radius 1.

The ellipse corresponding with a matrix $$ A =\ \left[\begin{matrix}a & b\\c &d\\\end{matrix}\right]= \ \left[\begin{matrix}1 & 0\\2 & 1\\\end{matrix}\right]$$ is defined by:

$$\left(d^2+c^2\right)x^2-2\left(ac+bd\right)xy+\left(a^2+b^2\right)y^2-\left(ad-bc\right)^2=0 $$ $$\left( 5\right)x^2-2\left(2\right)xy+\left( 1 \right)y^2-\left( 1 \right)^2=0 $$ or for the image of the circle with radius $2$: $$\left( 5\right)x^2-2\left(2\right)xy+\left( 1 \right)y^2-\left( 4 \right)=0 $$

The main axes of the ellipse are the singular vectors of the matrix and the length of the axes are the singular values.

If you replace the $1$ with $4$ in the reasoning below, you will find the exact equation you need. $4$ since the square of the norm is used in the calculations.

The image of the unit-circle by $A$ can be described as:
"the set of all $(u,v)$ that are the image of an $(x,y)$ on the unit circle"
$$\left\{\left[\begin{matrix}u\\v\\\end{matrix}\right]:\left[\begin{matrix}u\\v\\\end{matrix}\right]=A\left[\begin{matrix}x\\y\\\end{matrix}\right]\ and \ x^2+y^2=1\right\}$$

If we apply the inverse transformation on the vectors $(u,v)$ we must end up on the unit circle again:

$$\left\{{\left[\begin{matrix}u\\v\\\end{matrix}\right]:A}^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]=\left[\begin{matrix}x\\y\\\end{matrix}\right]\ and \ x^2+y^2=1\right\}$$

All $(u,v)$ on the ellipse are mapped back on the unit-circle by $A^{-1}$

$$\|A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right] \| =1$$ We now expand the norm of the vector $A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]$. The norm of a vector $\|w\|=\| w^T w \|$.

$$\left(\|A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]\|\right)^2=\left(A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]\right)^T\left(A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]\right)=1$$

$$\left[\begin{matrix}u\\v\\\end{matrix}\right]^T\left(A^T\right)^{-1}A^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]=1$$

$$\left[\begin{matrix}u\\v\\\end{matrix}\right]^T\left({AA}^T\right)^{-1}\left[\begin{matrix}u\\v\\\end{matrix}\right]=1$$

Expanding the expression above leads to the equation of the ellipse corresponding to a matrix $A$.

$$\left(d^2+c^2\right)u^2-2\left(ac+bd\right)uv+\left(a^2+b^2\right)v^2=\left(ad-bc\right)^2$$

Similar reasoning can be followed to calculate the ellipse corresponding to the inverse transform.
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