# Finding the angle value given 1 point and the centre of a circle

I got the coordinates of the center of a circle $(a,b)$ as well as one other point $(x, y)$. From those I can derive the radius by applying square root to the result of following formula.

$$(x-a)^2 + (y-b)^2 = r^2$$

This should allow me to compute the angle but I have only this formula:

$$x = a + r\cos t$$ $$y = b + r\sin t$$

How can I compute $t$?

I am a bit lost in solving this system of equations. Any help?

You know $x, y, a, b$ and $r$. Then $$\cos(t) = \frac{x-a}{r} = c_{1} \quad \text{and} \quad \sin(t) = \frac{y-b}{r} = c_{2}$$ Note that $c_{1}$ and $c_{2}$ are just two numbers that you compute based on what you know. How many solutions does each of the above trigonometric equations have in $t \in [0, 2\pi)$?
Of course, in the end, the right $t$ should satisfy both equations..
• Not really. For certain angles people usually know the cosine and sine values: for example if $t=30^{o}$ (or $t=\pi/6$ in radians), then we know $\cos{t} = \sqrt{3}/2$ and $\sin{t} = {1}/{2}$. So, inversely if $c_{2}=1/2$ in our example, then we know that $t=30^{o}$ or $t=30^{o} + 90^{0}=150^{o}$. Take a look at the unit circle for such angles: en.wikipedia.org/wiki/Trigonometric_functions#mediaviewer/… If $c_{1}$ and $c_{2}$ are arbitrary, you may use a scientific calculator to compute the inverse cos/sin (arccos/arcsin). Dec 21, 2014 at 16:34