Compute $\lim_{t\rightarrow0}\int_B \frac{l(t,x,y)}{t^2}d(x,y)$. This is a problem that I saw on the Internet. I'd like some hints to solve it.
Let $B$ be a ball in $\mathbb{R}^2$. Given $(x,y)\in B$, consider the circle of radius $t>0$ and centre $(x,y)$. Let $l(t,x,y)$ be the length of the arc of that circle that is outside of $B$. Find $$\lim_{t\rightarrow0}\int_B \frac{l(t,x,y)}{t^2}d(x,y).$$

Here is a picture: in black the boundary of the Ball $B$. In $\color{red}{\text{red}}$ (+ blue) the circle considered (centered at some $x,y$ with radius $t$). In $\color{blue}{\text{blue}}$ the arc whose length we define to be $l(t,x,y)$

 A: Update
The limit in question is $4\pi R$, where $R$ denotes the radius of the large disk. Here is a quick solution:

For fixed $t\ll R$ all the action is in the narrow zone $A:=\{(x,y)\>|\>R-t\leq\sqrt{x^2+y^2}\leq R\}$ immediately adjacent to the boundary circle  $\partial B$. Furthermore the tiny circles of radius $t$ cannot "feel" the curvature of the huge circle $\partial B$. We can therefore model $\partial B$ by the $y$-axis modulo $2\pi R$ and let the strip $A'\colon \>0\leq x\leq t$ model the  annulus  $A$. One then has
$$\ell(t,x,y)=2 t\,\phi= 2 t\>\arccos{x\over t}\qquad(0\leq x\leq t)$$
and therefore
$$\int_{A'}\ell(t,x,y)\>{\rm d}(x,y)=2\pi R\cdot 2 t\int_0^t\arccos{x\over t}\>dx=4\pi R t^2\int_0^1\arccos\tau\>d\tau=4\pi R t^2\ .$$
This proves the claim. The argument shows that the limit in question is $2L$ for any smooth curve of length $L$.
I'm appending the original solution which works with the "real" $B$:
Assume that the radius of the big disk $B$  is $1$, and fix a positive $t\ll1$. For a point $(x,y)\in B$ with $\sqrt{x^2+y^2}=:r$ the length $\ell(t,x,y)$ is given by
$\ell(t,x,y)=2t\phi$, whereby the involved data are related by
$$r^2+t^2+2rt\cos\phi=1\ .$$
We consider $\phi$ as primary variable. Then
$$r(\phi)=\sqrt{1-t^2\sin^2\phi}-t\cos\phi\ .$$
The relevant intervals are $0\leq\phi\leq{\pi\over2}+\arcsin{t\over2}$ and correspondingly $1-t\leq r(\phi)\leq1$ (draw a figure!). It follows that
$$\int_B\ell(t,x,y)\>{\rm d}(x,y)=2t\int_0^{\pi/2+\arcsin(t/2)}\phi\>2\pi r(\phi)r'(\phi)\>d\phi\ .$$
Expanding with respect to $t$  Mathematica computes
$$r(\phi)=1+?t,\qquad r'(\phi)=t(\sin\phi+?t)\ .$$
It follows that
$$\int_B\ell(t,x,y)\>{\rm d}(x,y)=4\pi t^2\left(\int_0^{\pi/2}\phi\sin\phi\>d\phi+?t\right)=4\pi t^2(1+?t)\ ,$$
hence the claim.
(I write $?t^k$ for the remainder terms Mathematica abbreviates with $O[t]^k$.)
A: Just a start, to show that the limit in question is positive, if it exists. For convenience I'll work in $B=B(0,1).$ Let $t>0,$ $t$ small. Suppose $1-t/2< r < 1.$ Then $r+t > 1 +t/2$ for all such $r.$ Some simple geometry then shows that $l(t,r,0) > t/2$ for all such $r.$ In the same way we have $ l(t,r\cos t,r\sin t) > t/2$ for $1-t/2< r < 1$ and any $t\in [0,2\pi].$ Therefore
$$\int_B l(t,x,y)\, dx\, dy = \int_0^{2\pi}\int_0^1 l(t,r\cos t,r\sin t) r\,dr\, dt \ge 2\pi\int_{1-t/2}^1 l(t,r\cos t,r\sin t) r\,dr \ge 2 \pi (t/2)(1-t/2)(t/2)\, dr,$$
which is on the order of $t^2.$
