A point is chosen at random from the interior of a circle. The circle has an equation of $x^2 + y^2 <= 961$. What is the density of this distribution at the origin, i.e. what is $f(0,0)$?
I would like to say I know where to even remotely start with the problem, but I do not. I am guessing I set up a double integral with an integration of $x^2 + y^2 dx$ dy involved. Any ideas would be greatly appreciated.
I read "A point is chosen at random from the interior of a circle" as a uniform distribution. You then mention that the circle is given by the equation $x^2 + y^2 \leq 961$, i.e. it is a circle with origin $0$ and radius $\sqrt{961}=31$.
Thus you should be interested in a uniform distribution on the set $B=\{(x,y)\in\mathbb{R^2} | x^2 + y^2 < 31^2\}$.
Now the density $f$ of a uniform distribution on a set $B$ satisfies (is defined such) that $f(x,y)=C \cdot \mathbb{1}_{B}(x,y)$, where $C$ is the constant that makes $f$ a probability density, i.e.
$\int_\mathbb{R^2} f(x,y) \, \mathrm{d}(x,y) = 1$.
Now
$\int_\mathbb{R^2} f(x,y) \, \mathrm{d}(x,y) = C \int_B \, \mathrm{d}(x,y)$,
such that $C$ is just the inverse of the area of the circle. Thus
$C=(\pi \cdot 31^2)^{-1} = (961\pi)^{-1}$,
and $f(0,0)=(961\pi)^{-1}$ as $(0,0)\in B$.