How to represent impulse function in 2D?
So dealing with generalized functions like the Dirac delta requires some care, and when dealing with N-dimensional versions you need to be very explicit with your notation to keep things straight.

I'll denote the 2 dimensional delta function in polar coordinates at the origin as ${}^2\delta(r, \theta) = {}^2\delta(r)$, since for the special case of the origin, $\theta$ doesn't matter.

For this derivation, ${}^2\delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):

$${}^2\delta(r) = \lim_{\epsilon \rightarrow 0^+} \dfrac{1}{\pi\epsilon^2} \quad 0 &lt; r &lt; \epsilon$$

Using this limiting sequence of functions, the 2-D delta function at the origin in polar coordinates has the following property of "integrating to 1" for the 2-D integration:

$$\int_0^{2\pi} \int_0^{\infty} {}^2\delta(r) \space r \space \mathrm{d}r \space \mathrm{d}\theta = 1$$

To separate the $r$ and $\theta$ portions of ${}^2\delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:

$${}^1\delta_a(r) = \lim_{\epsilon \rightarrow 0^+} \dfrac{1}{\epsilon} \quad 0 &lt; r &lt; \epsilon$$

Using this limiting sequence of functions, the 1-D delta function at the origin has the following property of "integrating to 1" for the 1-D integration:
$$\int_0^{\infty} {}^1\delta_a(r) \space \mathrm{d}r = 1$$

We can then equate the two above integrals to get the relationship between  $ {}^2\delta(r)$ and ${}^1\delta_a(r)$

$$\begin{align}\\
\int_0^{\infty} {}^1\delta_a(r) \space \mathrm{d}r &amp;= 1\\
&amp;= \int_0^{2\pi} \int_0^{\infty} {}^2\delta(r) \space r \space \mathrm{d}r \space \mathrm{d}\theta\\
&amp;= \int_0^{\infty} {}^2\delta(r) \space 2\pi r \space \mathrm{d}r\\
\end{align}$$

so we have

$${}^2\delta(r) = \dfrac{{}^1\delta_a(r)}{2\pi r}$$

This differs from your desired result by a factor of $2$, because I did not allow the 1 dimensional delta function to be symmertic around $0$, which wouldn't make sense for a polar origin where $r$ cannot be less than $0$ and $\theta$ ranges in $[0, 2\pi)$.

If one allows $r$ to go negative and $\theta$ to range in $[0, \pi)$, then one can show for a 1 dimensional delta function that is symmetric around $0$
$${}^2\delta(r) = \dfrac{{}^1\delta(r)}{\pi |r|}$$

(note the absolute value bars!)

This is a good example of why delta functions require careful definitions and considerations when making statements.

To show the projections of a 2-D delta function are themselves delta functions, start with a suitable limiting sequence of functions for ${}^2\delta(x,y)$, like an infinitely thin and tall rectangle function both in x and in y, such that

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} {}^2\delta(x,y) \space \mathrm{d}x \space \mathrm{d}y = 1$$

and the answer should fall out from the limiting sequence of functions you use.