The problem is from Gamelin exercise X.2.2
Assume that $u(x,y)$ is twice continuously differentiable function on a domain $D$. For $(x_0,y_0)\in D$, let $A_{\epsilon}(x_0,y_0)$ be the average of $u(x,y)$ on the circle centered at $(x_0,y_0)$ of radius $\epsilon$. Show that $$\lim_{\epsilon\to 0}{A_{\epsilon}(x_0,y_0) - u(x_0,y_0)\over\epsilon^2} = {1\over 4}\Delta u(x_0,y_0),$$ where $\Delta$ is the Laplacian operator
Hint is to use taylor expansion. Using the second order expansion, $$u(x,y) = u(x_0,y_0)+(x-x_0)u_x(x_0,y_0)+(y-y_0)u_y(x_0,y_0)+{1\over 2!}((x-x_0)^2u_{xx}(x_0,y_0)+2(x-x_0)(y-y_0)u_{xy}(x_0,y_0)+(y-y_0)^2u_{yy}(x_0,y_0)) + o(||(x-x_0,y-y_0)||^2).$$ Now, $A_\epsilon(x_0,y_0) = {1\over 2\pi}\int_0^{2\pi}u(z_0+\epsilon e^{i\theta}) d\theta$ where $z_0 = x_0+iy_0$. I think in the limit, enough to consider $(x-x_0)u_x(x_0,y_0)+\cdots+ (y-y_0)^2u_{yy}(x_0,y_0)$ term but no idea what to do next. Please help.
Without loss of generality we can assume that $(x_0, y_0) = (0, 0)$ and that $u(0, 0) = 0$.
Writing $U_x, U_y, U_{xx}, \ldots $ as a shorthand for $u_x(0, 0), u_y(0, 0), u_{xx}(0, 0), \ldots$ we have
$$ u(\epsilon \cos(t), \epsilon \sin(t)) = \epsilon \cos(t) U_x + \epsilon \cos(t) U_y \\ + \frac12 \left( \epsilon^2\cos^2(t)U_{xx} + 2\epsilon^2 \cos(t) \sin(t) U_{xy} + \epsilon^2 \sin^2(t)U_{yy}\right) + o(\epsilon^2) \, . $$
When this is integrated for $0 \le t \le 2 \pi$ then the terms involving $U_x, U_y, U_{xy}$ vanish, so that $$ A_\epsilon(0, 0) = \frac{1}{2 \pi} \left( \frac 12 \epsilon^2 U_{xx} \int_{0}^{2\pi} \cos^2(t) dt + \frac 12 \epsilon^2 U_{yy} \int_{0}^{2\pi} \sin^2(t) dt + o(\epsilon^2)\right) \, . $$ Both integrals evaluate to $\pi$, and we get $$ A_\epsilon(0, 0) = \epsilon^2 \left(\frac{U_{xx}+U_{yy}}{4} + o(1)\right) \, . $$ The desired conclusion now follows immediately.