# How to calculate new coordinates on a circle's circumference when an angle is given?

I am working on digital maps and I have a circle plotted. I have the circle's centre in lat, long and the circle's radius in meters.

Now I have a point on the north end (0 deg) of this circle's circumference. This is a web based application where the user will enter an angle in an input field and then I will take the value and move the point on the circle's circumference to that angle.

So for example a user enters 90, I'd like to calculate the coordinates to the east of the center, r meters away, where r is the radius.

Similarly if a user enters 180, I'd like to calculate the coordinates to the south of the center, r meters away, where r is the radius.

I know that 1 meter of lat, long is approximately $0.000009^\circ$ so I can convert it to degrees.

So basically I have the center and radius of a circle, and I will be provided with an angle. Can somebody tell me how to calculate the coordinates on the circumference on that particular angle assuming that the north most point is $0^\circ$.

• isn't $(r\sin t, r \cos t)$ is what you need?
– abel
Commented Jun 3, 2015 at 8:41
• I have no idea how to use that. Could you please elaborate? :) Commented Jun 3, 2015 at 8:42

In fact you know that $(x-x_0)^2+(y-y_0)^2=r^2$, where $(x_0,y_0)$ are the center coordinates and $r$ the radius, and $(x,y)$ the coordinates of the point you want.
The angle of the point relative to the $y$ axis is $\theta$.
If $\theta\neq \dfrac{\pi}{2}$, then you have
$\tan \theta=\dfrac{x-x_0}{y-y_0}$ or $(x-x_0)=\tan \theta (y-y_0)$
You then have $(y-y_0)^2(1+\tan^2 \theta)=r^2$
or $y=y_0 +\dfrac{\theta}{|\theta|}\dfrac{r^2}{1+\tan^2 \theta}$, and $x$ is defined also with $y$.