Assuming you are using a spherical model of the earth, a simple way is to proceed due north or due south, for then each degree of latitude has a constant length (with value extremely close to 10000000/90 meters, because the French revolutionaries originally defined the meter as 10^-7 times the distance from the equator to the North Pole along the Paris meridian!). For large radii or points close to poles you will have to adjust the latitude appropriately to account for passing over the pole (and add 180 degrees to the longitude).
For example, starting at (lon, lat) = (-90, 40) with a radius of 10,000 feet, you would convert that to 10,000 / (39.37/12) = 3048.006 meters, then convert that to 3048.006 * 90 / 10000000 = 0.0274321 degrees, landing you at (-90, 40.0274321). (Notice I write the longitude first, because it's the horizontal coordinate.)
If you need more accuracy, don't look up the approximate spherical radius of the earth: use the geoid appropriate for the datum of your coordinates. Appropriate formulae appear in books on geodesy and earth projections. A good resource is the USGS Professional Paper 1395, Map Projections--A Working Manual, by John Snyder.
Best Answer
This is a question of directional statistics. Similar issues arise with the average of circular quantities.
The conventional method is to convert latitudes ($\phi_i$) and longitudes ($\lambda_i$) into three dimensional points $(x_i,y_i,z_i)$ using
$$(R \cos \phi_i \cos \lambda_i, R \sin \phi_i \cos \lambda_i , R \sin \lambda_i )$$
then take the mean of these points $(\bar{x},\bar{y},\bar{z})$, generally giving you a point inside the sphere, and then converting this direction back to latitude and longitude, using something like $$\bar{\phi} = \text{atan2}\left(\bar{y},\bar{x}\right) \text{ and } \bar{\lambda} = \text{atan2}\left(\bar{z},\sqrt{\bar{x}^2+\bar{y}^2}\right). $$
Proportionately how far the mean point is inside the sphere, i.e. $\frac{\sqrt{\bar{x}^2+\bar{y}^2+\bar{z}^2}}{R}$, is an indicator of dispersion of the original points.