# Galactic distance equation

1. Jan 13, 2009

### Ayame17

I'm currently working on my final year project, and one of the little bits to do is to see if certain data points fall within a circle with my own defined radius and central co-ordinates. I've been given the equation to use:

$$d = \frac{\sqrt{(l-l_{0})^2 + (b-b_{0})^2}}{r}$$

where $$l_{0}$$ and $$b_{0}$$ are my central galactic co-ordinates, r is my radius (in degrees) and d is the distance the data point is from the central co-ordinates (also in degrees) - if d is less than 1, then the point is within my radius.

Although I don't need to, I'd just like to know where the equation comes from! I can see how the right hand side of the equals is a rearrangement of the equation of a circle, but I don't see how the distance is put in. Any help is appreciated!

2. Jan 13, 2009

### mgb_phys

3. Jan 13, 2009

### Ayame17

I realised that it must be Pythagoras, as that's where the equation for a circle comes from, it was more the fact that - when rearranged from the equation - the hypotenuese would be $$r^2d^2$$. Since it is normally just $$r^2$$, I wasn't sure if you were able to just put the distance ($$d^2$$) in.