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    <p>Assuming that your map is Google Maps, they use the Mercator projection, so you'd need to use that for the conversion. Under the Mercator projection, the distance that a pixel represents in meters varies with latitude, so while a meter is a very small distance compared to the Earth radius, latitude is important.</p> <p>All the examples below are javascript, so you might need to translate them.</p> <p>Here is a general explanation of the coordinate system:</p> <p><a href="http://code.google.com/apis/maps/documentation/javascript/maptypes.html#WorldCoordinates" rel="nofollow noreferrer">http://code.google.com/apis/maps/documentation/javascript/maptypes.html#WorldCoordinates</a></p> <p>This example contains a MercatorProjection object, which includes the methods fromLatLngToPoint() and fromPointToLatLng():</p> <p><a href="http://code.google.com/apis/maps/documentation/javascript/examples/map-coordinates.html" rel="nofollow noreferrer">http://code.google.com/apis/maps/documentation/javascript/examples/map-coordinates.html</a></p> <p>Once you have converted your (x,y) to (lat,lon), this is how you draw a circle:</p> <pre><code>// Pseudo code var d = radius/6378800; // 6378800 is Earth radius in meters var lat1 = (PI/180)* centerLat; var lng1 = (PI/180)* centerLng; // Go around a circle from 0 to 360 degrees, every 10 degrees for (var a = 0 ; a &lt; 361 ; a+=10 ) { var tc = (PI/180)*a; var y = asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc)); var dlng = atan2(sin(tc)*sin(d)*cos(lat1),cos(d)-sin(lat1)*sin(y)); var x = ((lng1-dlng+PI) % (2*PI)) - PI ; var lat = y*(180/PI); var lon = x*(180/PI); // Convert the lat and lon to pixel (x,y) } </code></pre> <p>These two mashups draw a circle of a given radius on the surface of the Earth:</p> <p><a href="http://maps.forum.nu/gm_sensitive_circle2.html" rel="nofollow noreferrer">http://maps.forum.nu/gm_sensitive_circle2.html</a></p> <p><a href="http://maps.forum.nu/gm_drag_polygon.html" rel="nofollow noreferrer">http://maps.forum.nu/gm_drag_polygon.html</a></p> <p>If you choose to ignore the projection then you'd use cartesian coordinates and simply draw the circle using Pythagoras Theorem:</p> <p><a href="http://en.wikipedia.org/wiki/Circle#Cartesian_coordinates" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Circle#Cartesian_coordinates</a></p>
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