$\lim F(a) = +\infty$ where $F(a):=\int_B \frac{1}{|z-a|} (1-|z|)^{-1/2}dxdy,\quad\forall a\in B.$

Let $B=\{(x,y)\in\mathbb{R}^2: x^2+y^2<1\}$. Let $z$ denote $(x,y)$, and let $$F(a):=\int_B \dfrac{1}{|z-a|} (1-|z|)^{-1/2}d xdy,\quad\forall a\in B.$$I want to verify that $\lim_{a\rightarrow \partial B }F(a) = +\infty$. Here $\partial B$ is the boundary of $B$.

Some facts that I knew are $$\int_B \dfrac{1}{|z-a|^{p}}dxdy <+\infty,\quad \forall a\in B,$$ if $1\leq p<2$, and $$\int_B (1-|z|)^{-p}dxdy =+\infty$$ if $p\geq 1$.

I saw that we only need to focus on $B^1:=\{z=(x,y):|z-a|>(1-|a|)/2 \}$ since over $B-B^1$ one has $1-|z| \geq |z-a|$, thus the integral over $B-B^1$ is uniformly bounded.

Any suggestion is helpful to me.