Let $ \delta_{x_0} $ be the distribution defined by $ \delta_{x_0} = \varphi (x_0) $, $ x_0 \in \mathbb{R^n} $. One can show that it has compact support, so in particular $ \delta \in \mathcal{S}'$, where $ \mathcal{S}' $ denotes the space of tempered distributions acting on the space of Schwartz functions $ \mathcal{S} $.
The Fourier transform of $ \delta $ is $$ \hat\delta_{x_0} (\varphi) = \delta_{x_0} (\hat\varphi) = \hat\varphi(x_0) = \int_\mathbb{R^n} e^{-i \langle x_0, y \rangle} \varphi(y) \: dy = \langle e^{-i \langle x_0, \cdot \rangle} \vert \varphi \rangle, \:\:\:\:\: \varphi\in \mathcal{S}, $$ which is well-defined since $ \lvert e^{-i \langle x_0, \cdot \rangle} \rvert \leq 1 $ and $ \mathcal{S} \in L^1 $, so that $$ \int_\mathbb{R^n} \lvert e^{-i \langle x_0, y \rangle} \varphi(y) \rvert \: dy \leq \int_\mathbb{R^n} \lvert \varphi(y) \rvert \: dy < +\infty. $$
In particular we find that $ \hat\delta = \hat\delta_0 = 1 $, i.e. $ \hat\delta (\varphi) = \int_\mathbb{R^n} \varphi(y) \: dy $. Please correct me if any part of my reasoning is false or unclear.
Now, I have come across the following statement and was wondering whether it is correct and, in case it is, what the meaning of it is and how to prove it:
$$ \hat\delta_{x-x_0} = \hat\delta_{x_0}. $$
I would be grateful for any help.