We can parameterize the circle by letting $z=z_0 + r e^{i\phi}$ Then $dz=ir e^{i\phi}d\phi$ Using the Cauchy integral formula we can express $f(z_0)$ in the following way: \begin{eqnarray*} f(z_0)&=&\frac{1}{2\pi i} \oint_{C} \frac{f(z)}{z-z_0}dz\\ &=&\frac{1}{2\pi i} \int_{0}^{2\pi} \frac{f(z_0 + r e^{i\phi})}{r e^{i\phi}} ir e^{i\phi} d\phi\\ &=&\frac{1}{2\pi} \int_{0}^{2\pi} f(z_0 + r e^{i\phi}) d\phi . \end{eqnarray*}