So here's what I got so far:
$r(t)$ is a circle iff $|r(t)| = \text{const.}$ iff $\sqrt{(a_1\cos(t) + b_1\sin(t))^2 + (a_2\cos(t) + b_2\sin(t))^2} = \text{const.}$
iff $$(a_1\cos(t) + b_1\sin(t))^2 + (a_2\cos(t) + b_2\sin(t))^2 = \text{const.}$$
iff $$a_1^2 \cos^2(t) + b_1^2\sin^2(t) + a_2^2\cos^2(t) + b_2^2\sin^2(t) + 2a_1b_1 \cos(t) \sin(t) + 2a_2b_2 \cos(t) \sin(t) = \text{const.}$$
iff $$a_1^2 \cos^2(t) + b_1^2\sin^2(t) + a_2^2\cos^2(t) + b_2^2\sin^2(t) + a_1b_1 \sin(2t) + a_2b_2 \sin(2t) = \text{const.}$$
iff $$a_1^2 \cos^2(t) + b_1^2\sin^2(t) + a_2^2\cos^2(t) + b_2^2\sin^2(t) + (a_1b_1 + a_2b_2) \sin(2t) = \text{const.}$$
iff $$(a_1^2 + a_2^2) \cos^2(t) + (b_1^2 + b_2^2)\sin^2(t) + (a_1b_1 + a_2b_2) \sin(2t) = \text{const.}$$
iff $$\langle a,a \rangle \cos^2(t) + \langle b,b \rangle\sin^2(t) + \langle a,b \rangle \sin(2t) = \text{const.}$$
iff $$\| a \|^2 \cos^2(t) + \| b \|^2 \sin^2(t) + \langle a,b \rangle \sin(2t) = \text{const.}$$
iff $$\| a \| \cos^2(t) + \| b \| \sin^2(t) + \langle a,b \rangle \sin(2t) = \text{const.}$$
It's easy for me to see that a sufficient condition is that the following two conditions hold
$$ \langle a,b \rangle = 0 $$ and $$ \| a \| = \| b \| $$
Now, did I do anything wrong up to this point? If not, I am really unsure if it's also a necessary condition and if there are any more sufficient and necessary conditions.
Let $A$ be the matrix with columns $a,b$. Then $r(t) = A \gamma(t)$, where $\gamma(t)=(\cos t, \sin t)$ is the canonical parametrization of the unit circle.
The image of the unit circle under a linear transformation is an ellipse. This is a consequence of the singular-value decomposition. Therefore, the image is a circle iff the singular values of $A$ are equal. The singular values of $A$ are the square roots of the eigenvalues of $AA^T$. Therefore, the image is a circle iff the eigenvalues of $AA^T$ are equal.
If $\langle a,b \rangle = 0$, then $AA^T$ is a diagonal matrix with diagonal entries $\| a \|^2$ and $\| b \|^2$. Hence your condition is sufficient.
More generally, the eigenvalues of $AA^T$ are equal iff $ (a_1^2 + a_2^2 + b_1^2 + b_2^2)^2 = 4(a_2 b_1 - a_1 b_2)^2 $.
This is an expression of the fact that $B$ has equal eigenvalues iff $(\operatorname{tr} B)^2 = 4 \det B$.