How to find the eigenvalues of the linear transformation that takes circle to ellipse

Question

I have the following question,

How to find the eigenvalues of the linear transformation $$T:\mathbb{R}^2\rightarrow \mathbb{R}^2$$ that takes the circle $$C=\{(x,y):x^2+y^2=1\}$$ to ellipse $$E=\{(x,y): {x^2\over 9} + {y^2 \over 4}=1\}$$ Options are

(1) $1,-1$

(2) $3,2$

(3) $-3,-2$

(4) None of the above

My Efforts

First we write the circle and ellipse in the parametric form, $$C=\{(\cos(t), \sin(t)): t \in (-\pi, \pi]\}$$ $$E=\{(3\cos(t), 2\sin(t)): t \in (-\pi, \pi]\}$$

But how do I know what $T$ is and how does it act on $\mathbb{R}^2$

Answer 1

There are infinitely many such $T$. One of them is diagonal and has eigenvalues $2,3$. Another is, as already mentioned in a comment, $T(x,y)=(-3y,2x)$, which has no real eigenvalues. So it's a bad question - if it actually asks about "the" $T$ that takes the circle to that ellipse it's a very bad question.