Determine limit of |a+b| This is a simple problem I am having a bit of trouble with. I am not sure where this leads.

Given that $\vec a = \begin{pmatrix}4\\-3\end{pmatrix}$ and $|\vec b|$ = 3, determine the limits between which $|\vec a + \vec b|$ must lie.

Let, $\vec b = \begin{pmatrix}\lambda\\\mu\end{pmatrix}$, such that $\lambda^2 + \mu^2 = 9$
Then, 
$$
\begin{align}
\vec a + \vec b &= \begin{pmatrix}4+\lambda\\-3 + \mu\end{pmatrix}\\
|\vec a + \vec b| &= \sqrt{(4+\lambda)^2 + (\mu - 3)^2}\\
&= \sqrt{\lambda^2 + \mu^2 + 8\lambda - 6\mu + 25}\\
&= \sqrt{8\lambda - 6\mu + 34}
\end{align}
$$
Then I assumed $8\lambda - 6\mu + 34 \ge 0$. This is as far I have gotten. I tried solving the inequality, but it doesn't have any real roots? Can you guys give me a hint? Thanks.

I would parameterize $b$ as $(3 \cos t, 3 \sin t)$, so the points to look at are $(4 + 3 \cos t, -3+ 3 \sin t)$. Find the value of $t$ that gives extreme values for the distance of this from the origin.
