Density of distance from center of circle to a chord

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Given two random points on the unit circle, A and B, define X to be the perpendicular distance from the center of the circle to the chord. Find the density function of X

My attempt: I was given a hint that because of symmetry we should fix one point and consider only the other point random. The definition of random was not given, so I fixed a point at $(1, 0)$ and let the other point be defined by its angle with the positive axis $\Theta\sim \text{Uniform}[0, 2\pi)$.

From this, I derived $$X=\left|\cos\left(\frac{\Theta}{2}\right)\right|$$ and manipulating $\mathbb{P}(X\leq x)$ and differentiating gave me a density function of $$f_{X}(x)=\frac{2}{\pi}\cdot\frac{1}{\sqrt{1-x^2}},\qquad 0\leq x < 1$$

However, this seems unintuitive and incorrect to me. Why would the density approach infinity as the distance $x$ approaches 1? Have I made a mistake somewhere?

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Your answer is okay.

For $x\in[0,1]$ we find: $$P(X\leq x)=P\left(-x\leq\cos\left(\frac{\Theta}2\right)\leq x\right)=P\left(\arccos(x)\leq\frac{\Theta}2\leq\arccos(-x)\right)=$$$$\frac1{\pi}(\arccos(-x)-\arccos(x))=\frac1{\pi}(\pi-2\arccos(x))=1-\frac2{\pi}\arccos(x)$$

Taking the derivative you found correctly the density of $X$.

"Why would the density approach infinity as the distance $x$ approaches 1?"

Well, why not?

Notice that $\frac{P(1-\epsilon<X\leq 1)}{\epsilon}\to+\infty$ if $\epsilon\downarrow0$ which is in accordance with a density approaching infinity.