I answered part a, and then separated my answer into the real and imaginary parts in part b, and I don't know what I could do from there. The mark scheme for this question just substitutes this into the equation of a circle, which gets a value for k. I just don't see how that shows that the circle C has the equation in the question. Any help would be much appreciated. Thanks!
Well, since you solved part $a)$, you found that the constants are $a=-32, b=40, c=8$. After separating $w$ in real and imaginary parts, you want to find a point $p=(c_1,c_2)$ in the $w$ plane such that $\lvert w-p \rvert=k^2$, for $w$ as in the formula in part $a)$, which is already only for $z=x+ix$ in the $z$ plane; i.e., you want to solve for $(c_1,c_2)$ that represents where the center of the circle is. It is easier first to verify that the point $(3,0)$ is what you want. You will end up wanting to show that $$ \left( \frac{-32x^2+8}{16x^2+1}-3 \right)^2 + \left( \frac{40x}{16x^2+1}\right)^2=k^2$$ for some constant $k^2$. Developing the algebra a little bit, you'll reach the following polynomial division:$$\frac{6400x^4+800x^2+25}{(16x^2+1)^2}=25$$ which is constant! So $(3,0)$ is the center of the desired circle. If you wanted to discover that $(3,0)$ should be the center, it would be essentially the same logic, although with a lot more calculation.