Where does the galactic distance equation come from?

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SUMMARY

The galactic distance equation, defined as d = √((l - l₀)² + (b - b₀)²) / r, is derived from the Pythagorean theorem, which calculates the distance from a point to a central coordinate in a two-dimensional space. In this context, l₀ and b₀ represent the central galactic coordinates, while r is the defined radius in degrees. The equation effectively determines if a data point falls within a specified radius by comparing the calculated distance d to the radius r. The discussion clarifies that the equation is a rearrangement of the standard equation of a circle.

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Ayame17
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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:

[tex]d = \frac{\sqrt{(l-l_{0})^2 + (b-b_{0})^2}}{r}[/tex]

where [tex]l_{0}[/tex] and [tex]b_{0}[/tex] 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!
 
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I realized 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 [tex]r^2d^2[/tex]. Since it is normally just [tex]r^2[/tex], I wasn't sure if you were able to just put the distance ([tex]d^2[/tex]) in.
 

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