Variance of a Monte Carlo quarter-circle pi estimate

To estimate $\pi$ you draw $N$ independent uniform points in $[0,1]^2$. Let $I_i=1$ if the $i$th point satisfies $x^2+y^2\le1$ and $0$ otherwise, and let $T_N=\tfrac1N\sum_{i=1}^N I_i$. Using $T_N$ to build your estimate of $\pi$ (note $T_N$ by itself does not estimate $\pi$), the variance of that estimate has the form $\dfrac{a\pi^2+b\pi}{N}$ with integers $a,b$. Find $a+b$.

$3$. $\mathbb{E}[I_i]=\pi/4$, so the natural estimator is $4T_N$. Its variance is $\tfrac{16}{N}\cdot\tfrac{\pi}{4}\left(1-\tfrac{\pi}{4}\right)=\tfrac{4\pi-\pi^2}{N}$, giving $a=-1$, $b=4$.