Finding the affine transformation that turns the unit circle into $x^2+y^2+xy-x-y =1$ 
How to find an affine transformation $T: \mathbb{R^2} \to \mathbb{R^2}$ such that
$$T(\lbrace (x,y) \in \mathbb{R^2}:x^2+y^2 \leq 1\rbrace)= \lbrace (x,y) \in \mathbb{R^2}:x^2+y^2+xy-x-y \leq 1 \rbrace)$$

The affine transformation can be computed with $\vec{y}=A \vec{x}+b$.
Since it's the unit circle I tried to map it:
$$T((0,0)^T)=A(0,0)^T+b=(0,0)+b$$
but I don't get the transformation.
In which way can it be done?
 A: Let's first picture the set on the right:

Now that we know what we are dealing with an ellipse that is tilted by $45^\circ$ it is natural to try to rewrite the latter using the coordinates $t=x+y$ and $u=x-y$. Now since it's an ellipse it makes sense to try rewriting
$$x^2+y^2+xy-x-y=a\cdot(x+y+b)^2+c\cdot(x-y+d)^2-ab^2-cd^2.$$
Matching coefficients you get the conditions
\begin{align}a+c=1\\
2a-2c=1\\
2ab+2cd=-1\\
2ab-2cd=-1
\end{align}
solving these you get everything you need. Indeed you get
$$a=\frac34,\ c=\frac14,\ b=-\frac{2}3,\ d=0$$
So you can rewrite the equation as
\begin{align}x^2+y^2+xy-x-y&=\frac34\cdot(x+y-2/3)^2+\frac14\cdot(x-y)^2-1/3\\ &=\left(\frac{\sqrt{3}}{2}(x+y)-\frac{\sqrt{3}}3\right)^2+\left(\frac{x-y}{2}\right)^2-1/3.\end{align}
Now we go from
$$\left(\frac{\sqrt{3}}{2}(x+y)-\frac{\sqrt{3}}3\right)^2+\left(\frac{x-y}{2}\right)^2-1/3\leq 1$$
to
$$\frac34\left(\frac{\sqrt{3}}{2}(x+y)-\frac{\sqrt{3}}3\right)^2+\frac34\left(\frac{x-y}{2}\right)^2\leq 1$$
which is equal to
$$\left(\frac{3}{4}(x+y)-\frac12\right)^2+\left(\frac{\sqrt{3}}{4}(x-y)\right)^2\leq 1.$$
So the affine transform is
$$\begin{pmatrix}\frac{3}{4} & \frac{3}{4}\\ \frac{\sqrt{3}}{4} & -\frac{\sqrt{3}}{4}\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}+\begin{pmatrix}-\frac12 \\ 0 \end{pmatrix}$$
