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$x_a'$ and $y_a'$ are unknown. What's the simplest way to solve it? Every time I tried, it grew into tremendous size or was unable to think out in reasonable amount of time due to it's complexity.

$\begin{align*} (x_f-x_a')^2+(y_f-y_a')^2&=r^2\\ (x_a-x_a')^2+(y_a-y_a')^2&=4r^2\sin^2(\frac12\alpha) \end{align*}$

P.S. I don't need to specify real value, that is, I want to get a function $x_a'$ and $y_a'$ of known values.

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You can start by finding the radical line of your two circles. See this as well. – J. M. Jul 13 '12 at 18:20
up vote 2 down vote accepted

Expand the squares on the left-hand sides and subtract the two equations. The $x_a'^2$ and $y_a'^2$ terms will cancel out, and what is left is a linear equation in $x_a'$ and $y_a'$ which you can use to express $y_a'$ in terms of $x_a'$ (or vice versa -- one of these may be impossible depending on the constants).

Stick the expression for $y_a'$ into one of the original equations and simplify. The result is a quadratic equation in $x_a'$ which you can solve with the quadratic formula. Finally use the linear equation again to find the corresponding value of $y_'a$.

If you try to unfold this as a single closed formula in $x_f$, $y_f$, $r$ $x_a$, $y_a$, and $\alpha$, the resulting formula is going to be huge alright. But if you give names to some of the intermediate values and split the whole thing into steps, each step is individually simple enough.

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"...and what is left is a linear equation." - and that's the (equation of the) radical line of those two circles. – J. M. Jul 13 '12 at 18:25

In the plane, your equations describe three points $$ F=(x_f, y_f),\quad A=(x_a, y_a),\quad A'=(x_{a'},y_{a'}) $$ Where $F\text{ and }A$ are known, $A'$ is what you're seeking. You know the distances $| FA| = r^2$ and $|AA'| = 4r^2\sin^2(\alpha/2),$ which is fortunate, since a judicious use of the half-angle formula $$ 2\;\sin^2(\alpha/2)=1-\cos\alpha $$ and the law of cosines gives you enough information to conclude that the angle $\:\angle AFA' = \alpha.$ Now that you know that, a bit of translation and rotation will give you $A'$. The resulting solution isn't any prettier than what you'd get by using Henning's solution, sorry to say. As my thesis advisor once said to me, "sometimes things are just plain ugly."

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