This is a famous problem known as Gull's Lighthouse, from an example by Gull in 1988. It has deep implications when taken one additional step in both the social sciences and in physics. You actually have enough information to solve this problem through acceptance-rejection testing, but if you want to use MCMC feel free.

Let's look at your problem another way, knowing that the Lighthouse is 100 m from shore is actually a lot of information. That piece is a big deal, otherwise you would have to solve both distance and location.

As a side note, the Cauchy distribution is 636 semi-interquartile ranges wide at 99.99% HDR so if you were looking for narrow intervals, you can almost forget it. If you use a normal approximation, however, it will never converge as this problem violates the central limit theorem and if you estimate $\sigma$ as standard deviation then if $n$ is sample size then $s^2\to\infty$ as $n\to\infty$. If you did that, your estimates would get worse and worse as you added date points.

You need to note that you are observing the set $\{x_i,i\in{1\dots{n}}\}$ and there is some point $\mu$ where the lighthouse would be perpendicular to the beach. Note that it is not the positioning of the $x_i$ which are uniformly distributed, but rather the angle, $\theta$. We did not observe the angle, only the set of points.

The relationship of the points to the angle is $$\frac{x_i-\mu}{100}=\tan\theta.$$ In terms of the angle, this can be expressed as $$\theta=\tan^{-1}\left(\frac{x_i-\mu}{100}\right)$$

Since $$p(x_i|\mu,100)=p(\theta|\mu,100)|\frac{\mathrm{d}\theta}{\mathrm{d}x_i}|$$ and $\theta$ is uniform over the region $(-\pi/2,\pi/2)$ we know that $$p(x_i|\mu,100)=\frac{1}{\pi}|\frac{\mathrm{d}\theta}{\mathrm{d}x_i}|,$$ which is $$p(x_i|\mu,100)=\frac{1}{\pi}\frac{100}{100^2+(x_i-\mu)^2}.$$

So, by Bayes theorem, if the shore point perpendicular to the lighthouse is $(\mu,0)$, then the lighthouse is located at $(\mu,100)$. The estimator of this point, since you have no prior information on $\mu$ is $$p(\mu|x_1\dots{x_n},100)\propto\prod_{i=1}^n\frac{1}{\pi}\frac{100}{100^2+(x_i-\mu)^2},\forall\mu\in\Re.$$

If you want to make the problem more interesting, assume you do not know that the location is one hundred meters from shore. Your problem them becomes the two dimensional problem more suited to stan. Your problem then becomes $$p(\mu,\gamma|x_1\dots{x_n})\propto\prod_{i=1}^n\frac{1}{\pi}\frac{\gamma}{\gamma^2+(x_i-\mu)^2},\forall\mu\in\Re;\gamma\in\Re^{++}.$$

This is a famous problem known as Gull's Lighthouse, from an example by Gull in 1988. It has deep implications when taken one additional step in both the social sciences and in physics. You actually have enough information to solve this problem through acceptance-rejection testing, but if you want to use MCMC feel free.

Let's look at your problem another way, knowing that the Lighthouse is 100 m from shore is actually a lot of information. That piece is a big deal, otherwise you would have to solve both distance and location.

As a side note, the Cauchy distribution is 636 semi-interquartile ranges wide at 99.99% HDR so if you were looking for narrow intervals, you can almost forget it. If you use a normal approximation, however, it will never converge as this problem violates the central limit theorem and if you estimate $\sigma$ as standard deviation then if $n$ is sample size then $s^2\to\infty$ as $n\to\infty$. If you did that, your estimates would get worse and worse as you added date points.

You need to note that you are observing the set $\{x_i,i\in{1\dots{n}}\}$ and there is some point $\mu$ where the lighthouse would be perpendicular to the beach. Note that it is not the positioning of the $x_i$ which are uniformly distributed, but rather the angle, $\theta$. We did not observe the angle, only the set of points.

The relationship of the points to the angle is $$\frac{x_i-\mu}{100}=\tan\theta.$$ In terms of the angle, this can be expressed as $$\theta=\tan^{-1}\left(\frac{x_i-\mu}{100}\right)$$

Since $$p(x_i|\mu,100)=p(\theta|\mu,100)|\frac{\mathrm{d}\theta}{\mathrm{d}x_i}|$$ and $\theta$ is uniform over the region $(-\pi/2,\pi/2)$ we know that $$p(x_i|\mu,100)=\frac{1}{\pi}|\frac{\mathrm{d}\theta}{\mathrm{d}x_i}|,$$ which is $$p(x_i|\mu,100)=\frac{1}{\pi}\frac{100}{100^2+(x_i-\mu)^2}.$$

So, by Bayes theorem, if the shore point perpendicular to the lighthouse is $(\mu,0)$, then the lighthouse is located at $(\mu,100)$. The estimator of this point, since you have no prior information on $\mu$ is $$p(\mu|x_1\dots{x_n},100)\propto\prod_{i=1}^n\frac{1}{\pi}\frac{100}{100^2+(x_i-\mu)^2},\forall\mu\in\Re.$$

If you want to make the problem more interesting, assume you do not know that the location is one hundred meters from shore. Your problem them becomes the two dimensional problem more suited to stan. Your problem then becomes $$p(\mu,\gamma|x_1\dots{x_n})\propto\prod_{i=1}^n\frac{1}{\pi}\frac{\gamma}{\gamma^2+(x_i-\mu)^2},\forall\mu\in\Re;\gamma\in\Re^{++}.$$

As a side note, this problem is related to the also famous Witch of Agnesi, studied by Maria Agnesi and published in 1748 and the more basic problem studied by Fermat and Grandi in 1703. She had included it in the very first true calculus textbook. It provided students an education from algebra through integral and differential equations.