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Get radius from a circle to screen corners

I followed an online tutorial, I did not understand how the professor managed to get the distance of a circle from the top left corner of the screen:
//calculate the distance from epicenter (of a circle) to the top left corner of the screen
double theta = atan(epicenter.dy/epicenter.dx);
double distanceToCorner = epicenter.dy / sin(theta);
I would like to know how to get the distance from all the other screen corners (and possibly have an explanation of what has been done).
Thank you

Assuming, that in android you can get screen width and height, you can simply count the distance at horizontal and vertical axis separately.
Getting the distances at those axes, you can use Pythagoras equation, like
dist = sqrt( dx^2 * dy^2 )
To make it more felxible, just make a function, that takes corner position as a parameter and make the dx and dy as a absolute difference of corner and epicenter location.
Going back to your question and atan(...), I don't quite understand the need of using this here, except if that's a project for the math class :)

I know this is not the answer but if i understand what you mean, then this image might be helpful.

Java calculating points of Angles in a non-right triangle

I am currently working on a project in which i need to draw a non-right triangle in the center of a JFrame using either, java.awt.Graphics's drawLine() or drawPolygon() methods. Both of these methods require the coordinates of all of the points to function. My problem is that instead of points, all i have are all of the angles and side lengths of the triangle. I've drawn up a nifty diagram of what I hope helps you visualize my problem:
(EDIT the position of C in this Senario is not fixed betwen a and b and may be anywhere below the axis on which AB rests)
as you can see from my picture, I need the coordantes of C based off the coordanes of A, is there any way to calculate this given the lengths of all sides and angles of the non-right triangle?
Bonus: How would i find an (x, y) value for A that would effectivly center the triangle in the middle of the JFrame?

If you know angle CAB, the coordinate of point C should be:
(x+b·sin(θ), y-b·cos(θ))
In Java, there is:
double Math.sin(double radians);
double Math.cos(double radians);
Keep in mind that the angle needs to be in radians. If your angles are in degrees, try:
double Math.sin(Math.toRadians(double degrees));
double Math.cos(Math.toRadians(double degrees));
Hope this helps.

How to map 3D coordinates to 2D plane

I'm writing a OpenGL application in which there is a rectangle placed on a 3D object. The 3D object can move around and rotate and the rectangle follows these movements.
What I would like to do is to point with the mouse towards the rectangle so that a dot would appear in that point on the rectangle, and I want the point to follow it as the 3D object that "holds" the rectangle moves around.
I know how to find the intersection on a plane, and I know how to find the world coordinates of the contact point. What I need is a way to convert the world coordinates to the 2D local coordinate system of the rectangle.
For example, suppose I have the plane positioned like this in the 3D world, with a given orientation (sorry, I can't really draw properly):
The black point in the center is the origin of the plane, while the blue point is the one I would like to find. The numbers near the points are their world coordinates; in this example the Z axis "comes out" of the screen.
I would like to map the coordinates of the blue point in the local coordinate system of the plane, like this:
I know that somehow this shouldn't be hard, but I can't find a way at all. Any hints would be appreciated!

You probably already use a transformation matrix to render the rectangle. If you have the 3D position of the point, just transform it with the inverse transformation matrix and you get a 3D position in the rectangle's space. If the local system is aligned with the rectangle, one of the coordinates should always be zero and you can drop it.

I found what I needed in this post here.
Basically the solution for me is to project the point onto the x and y axis of the local coordinate system of the rectangle.
In pseudo code it's like this:
// I compute the direction that go from the origin of the rectangle to its x axis, in world coordinates, and take the norm; same thing for the y axis
var xDir = Norm(mRectangle.LocalToWorld([1, 0, 0]) - mRectangle.LocalToWorld([0, 0, 0]));
var yDir = Norm(mRectangle.LocalToWorld([0, 1, 0]) - mRectangle.LocalToWorld([0, 0, 0]));
// I compute the dot product
mLocalPosition.x = Dot(xDir, (contactPoint - mRectangle.LocalToWorld([0, 0, 0])));
mLocalPosition.y = Dot(yDir, (contactPoint - mRectangle.LocalToWorld([0, 0, 0])));
// I can now set the position of the point converting the local coordinates found to world coordinates
mPoint.SetPosition(mRectangle.LocalToWorld([mLocalPosition.x, mLocalPosition.y, 0]));

Calculating visible radius in Google Maps on Android

I'm trying to figure out how to calculate the largest radius of the screen's visible region.
(Before you say: Hey! Rectangulars don't have a radius! I understand that screen is rectangular (if camera is positioned in 90 degrees and not tilted), I'm talking about the radius of an imaginary circle that wraps the screen, or, the distance of the edge between the center of the screen to one of the corners of the rectangular, which is the same).
So, I understand that in a normal situation when screen is not tilted, I could take the distance between VisibleRegion.latLngbounds.southwest and the center of the screen to find the radius, but when screen is tilted, it becomes trapezoid.
Now, if the screen is an isosceled trapezoid, then I could take VisibleRegion.farLeft for example (which would be equal to farEast) and calculate the distance this way:
VisibleRegion vr = map.getProjection().getVisibleRegion();
Location center = new Location("center");
center.setLatitude(screenCenter.latitude);
center.setLongitude(screenCenter.longitude);
Location farVisiblePoint = new Location("farPoint");
farVisiblePoint.setLatitude(vr.farLeft.latitude);
farVisiblePoint.setLongitude(vr.farLeft.longitude);
float radius = center.distanceTo(farVisiblePoint);
My question is:
Is there any situation with the map that the calculation above will be wrong? Can the trapezoid not be an isosceled one?

Finding angle from center point on circle [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Find angle of a point from center of circle
Imagine a circle, the center point is fixed, as is the point at the top of the circle. What i want is to give a third point anywhere around the circle (or outside the circle) and get the degrees from the center to top point line from 0-359. (I actually graphed out a nice picture illustrating but im new and cant post it)
To give some context, the circle is a little planet and I want do place a little building on the surface with the base of the building at a tangent. I need the rotation angle for my building bitmap.
edit: Thanks for the help, i'm still struggling with this one though. I wonder could it be relevant that I'm using android and the y0 coordinate is at the top? Is it the other way around on other platforms? would that affect the calculation?
Solution: Because I am in android and the y coords are counted from top to bottom I had to change a - witha +
degrees = Math.atan2(x - centerX, -y + centerY);
// and to make it count 0-360
if (degrees < 0) {degrees += 2 * Math.PI;}

Use Math.atan2() to get the angle in radians from east, and then rotate and convert as appropriate.
Note that atan2 is defined as atan2(y, x) NOT atan2(x, y), as you might expect.

Get the horizontal distance and the vertical difference between the center and the point, divide one by the other, and pass the result to the method Math.asin(double).
The result will be the angle in radians. When you need it in degree, you can use the method Math.toDegrees(double). Most APIs I know prefer radians, though.