What is the radius and center of the image of $|z|=1$ under $ f(z) = \frac{3z+2}{4z+3}$?

Here is an automatic procedure: first invert the relation $w=f(z)$, then apply the condition $|z|=1$ to the inverse formula $z=g(w)$ to deduce an equation of the image set.

In the present case, $w=f(z)$ means that $$w=\frac{3z+2}{4z+3}$$ that is, $(4z+3)w=3z+2$, that is, $(4w-3)z=2-3w$, that is, $$z=\frac{2-3w}{4w-3}$$ Thus, the image of the circle has equation $$\left|\frac{2-3w}{4w-3}\right|=1$$ In turn, this means successively that $$|2-3w|=|4w-3|$$ that is, $$|2-3w|^2=|4w-3|^2$$ that is, $$4-6(w+\bar w)+9|w|^2=16|w|^2-12(w+\bar w)+9$$ that is, $$7|w|^2-6(w+\bar w)+5=0$$ and finally, if $w=x+iy$, $$7(x^2+y^2)-12x+5=0$$ from which you might be able to conclude that the desired radius is $$r=\frac17$$ As one can see, switching to the decomposition of complex numbers into their real and imaginary parts as late as possible in the computations, simplifies these.

Edit: The comment by user @alex.jordan below shows eloquently that "as late as possible" just above, could even be replaced by "never"...

Let see where this transformation takes $1,-1$ and $i$:

\begin{eqnarray} 1&\longmapsto &{5\over 7}\\-1&\longmapsto &1\\i&\longmapsto &{18+i\over 25}\\ \end{eqnarray}

Now calculate the center and radius of a triangle on $\alpha ={5\over 7}$, $\beta =1$ and $\gamma ={18+i\over 25}$.

Since this triangle is right at $\gamma$ we see that midpoint of segment $\alpha \beta$, that is $\sigma = {6\over 7}$ is a center of new circle with $r = {1\over 7}$.