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Convert 2 vector2 points to a rectangle in xna/monogame

I have some code which will detect the start and end point of a click-and-drag action, and will save it to 2 vector2 points. I then use this code to convert:
public Rectangle toRect(Vector2 a, Vector2 b)
{
return new Rectangle((int)a.X, (int)a.Y, (int)(b.X - a.X), (int)(b.Y - a.Y));
}
The code above does not work and googling, so far has come up inconclusive.
Could anyone please provide me with some code or a formula to properly convert this?
Note: a vector2 has an x and a y, and a rectangle has an x, a y, a width, and a height.
Any help is appreciated! Thanks

I think you need to have additional logic in there to decide which vector to use as the top left and which to use as the bottom right.
Try this:
public Rectangle toRect(Vector2 a, Vector2 b)
{
//we need to figure out the top left and bottom right coordinates
//we need to account for the fact that a and b could be any two opposite points of a rectangle, not always coming into this method as topleft and bottomright already.
int smallestX = (int)Math.Min(a.X, b.X); //Smallest X
int smallestY = (int)Math.Min(a.Y, b.Y); //Smallest Y
int largestX = (int)Math.Max(a.X, b.X); //Largest X
int largestY = (int)Math.Max(a.Y, b.Y); //Largest Y
//calc the width and height
int width = largestX - smallestX;
int height = largestY - smallestY;
//assuming Y is small at the top of screen
return new Rectangle(smallestX, smallestY, width, height);
}

EMGU CV MCvBox2D.GetVertices gives different order of corners

I'm using EMGU CV for my project and I'm facing a weird problem.
I use cannyedges to find some squares in a photo.
This is working correctly. After that i want to take a pixel inside each square and use that to draw the border.
The problem I'm facing is that i need the vertices of each corner in order to generate a random pixel.
To do this I use the code:
PointF[] corners = rectangle.GetVertices();
float x = Math.Max(corners[1].X, corners[0].X);
float y = Math.Max(corners[1].Y, corners[2].Y);
float width = Math.Min(corners[2].X, corners[3].X) - x;
float height = Math.Min(corners[0].Y, corners[3].Y) - y;
The problem with this code is that rectangle.GetVertices(); gives a different order of corners each time.
The first rectangle returns bottomleft as corner 0, top left as corner 1 etc.
How ever the second rectangle returns bottomright as corner 0, bottomleft as corner 1 etc.
I'm wondering if anyone else is having this problem and if anyone knows how to fix this?
If you need more info to answer this problem please tell me.

PointF[] corners = rectangle.GetVertices();
// Maybe this
corners = corners.OrderBy(s => s.X).ThenBy(s => s.Y).ToArray();
float x = Math.Max(corners[1].X, corners[0].X);
float y = Math.Max(corners[1].Y, corners[2].Y);
float width = Math.Min(corners[2].X, corners[3].X) - x;
float height = Math.Min(corners[0].Y, corners[3].Y) - y;

Find the point on a circle with given center point, radius, and degree

It's been 10 years since I did any math like this... I am programming a game in 2D and moving a player around. As I move the player around I am trying to calculate the point on a circle 200 pixels away from the player position given a positive OR negative angle(degree) between -360 to 360. The screen is 1280x720 with 0,0 being the center point of the screen. The player moves around this entire Cartesian coordinate system. The point I am trying trying to find can be off screen.
I tried the formulas on article Find the point with radius and angle but I don't believe I am understanding what "Angle" is because I am getting weird results when I pass Angle as -360 to 360 into a Cos(angle) or Sin(angle).
So for example I have...
1280x720 on a Cartesian plane
Center Point (the position of player):
let x = a number between minimum -640 to maximum 640
let y = a number between minimum -360 to maximum 360
Radius of Circle around the player: let r always = 200
Angle: let a = a number given between -360 to 360 (allow negative to point downward or positive to point upward so -10 and 350 would give same answer)
What is the formula to return X on the circle?
What is the formula to return Y on the circle?

The simple equations from your link give the X and Y coordinates of the point on the circle relative to the center of the circle.
X = r * cosine(angle)
Y = r * sine(angle)
This tells you how far the point is offset from the center of the circle. Since you have the coordinates of the center (Cx, Cy), simply add the calculated offset.
The coordinates of the point on the circle are:
X = Cx + (r * cosine(angle))
Y = Cy + (r * sine(angle))

You should post the code you are using. That would help identify the problem exactly.
However, since you mentioned measuring your angle in terms of -360 to 360, you are probably using the incorrect units for your math library. Most implementations of trigonometry functions use radians for their input. And if you use degrees instead...your answers will be weirdly wrong.
x_oncircle = x_origin + 200 * cos (degrees * pi / 180)
y_oncircle = y_origin + 200 * sin (degrees * pi / 180)
Note that you might also run into circumstance where the quadrant is not what you'd expect. This can fixed by carefully selecting where angle zero is, or by manually checking the quadrant you expect and applying your own signs to the result values.

I highly suggest using matrices for this type of manipulations. It is the most generic approach, see example below:
// The center point of rotation
var centerPoint = new Point(0, 0);
// Factory method creating the matrix
var matrix = new RotateTransform(angleInDegrees, centerPoint.X, centerPoint.Y).Value;
// The point to rotate
var point = new Point(100, 0);
// Applying the transform that results in a rotated point
Point rotated = Point.Multiply(point, matrix);
Side note, the convention is to measure the angle counter clockwise starting form (positive) X-axis

I am getting weird results when I pass Angle as -360 to 360 into a Cos(angle) or Sin(angle).
I think the reason your attempt did not work is that you were passing angles in degrees. The sin and cos trigonometric functions expect angles expressed in radians, so the numbers should be from 0 to 2*M_PI. For d degrees you pass M_PI*d/180.0. M_PI is a constant defined in math.h header.

I also needed this to form the movement of the hands of a clock in code. I tried several formulas but they didn't work, so this is what I came up with:
motion - clockwise
points - every 6 degrees (because 360 degrees divided by 60 minuites is 6 degrees)
hand length - 65 pixels
center - x=75,y=75
So the formula would be
x=Cx+(r*cos(d/(180/PI))
y=Cy+(r*sin(d/(180/PI))
where x and y are the points on the circumference of a circle, Cx and Cy are the x,y coordinates of the center, r is the radius, and d is the amount of degrees.

Here is the c# implementation. The method will return the circular points which takes radius, center and angle interval as parameter. Angle is passed as Radian.
public static List<PointF> getCircularPoints(double radius, PointF center, double angleInterval)
{
List<PointF> points = new List<PointF>();
for (double interval = angleInterval; interval < 2 * Math.PI; interval += angleInterval)
{
double X = center.X + (radius * Math.Cos(interval));
double Y = center.Y + (radius * Math.Sin(interval));
points.Add(new PointF((float)X, (float)Y));
}
return points;
}
and the calling example:
List<PointF> LEPoints = getCircularPoints(10.0f, new PointF(100.0f, 100.0f), Math.PI / 6.0f);

The answer should be exactly opposite.
X = Xc + rSin(angle)
Y = Yc + rCos(angle)
where Xc and Yc are circle's center coordinates and r is the radius.

Recommend:
public static Vector3 RotatePointAroundPivot(Vector3 point, Vector3 pivot, Vector3 angles)
{
return Quaternion.Euler(angles) * (point - pivot) + pivot;
}

You can use this:
Equation of circle
where
(x-k)2+(y-v)2=R2
where k and v is constant and R is radius

Calculate the coordinates in a circle

I am drawing an imaginary circle around the middle of a button.
The radius of the circle is the Height/2 if Height>Width or Width/2 if Width>Height.
Now i have to calculate which coordinates (in pixels) are in this circle.
The idea is that if e.g. the mouse cursor hovers over that circle, something happens.

Calculating each co-ordinate is overkill; just compare the distance to the center. For example:
int radius = 5; // whatever
int deltaX = originX - mouseX, deltaY = originY - mouseY;
// compare the square distance, to avoid an unnecessary square-root
if((deltaX * deltaX) + (deltaY * deltaY) <= (radius * radius)) {
// inside the circle, or on the edge
}
To avoid a little math, you could also do a quick bounding-box check, i.e. checking the rectangular region (just addition/subtraction). This can be used in combination, i.e.
check the bounding box
if it isn't in the bounding box it certainly isn't in the circle
if it is in the bounding box, do the math to compare the square-distance

You are inside the circle when this equation is satisfied:
Math.pow(mouse_pos_x-center_circle_x,2)+Math.pow(mouse_pos_y-center_circle_y,2)<Math.pow(radius,2)

The area of a a circle by definition is a group of points whose distance is equal to or less than the center.
All you must do to test if a point is within a circle is to calculate the distance between it and the center point. If this distance is smaller than the radius of the circle, the point is within the circle.
double Distance(Point p1, Point p2)
{
int x = p1.X - p2.X;
int y = p1.Y - p2.Y;
return Math.Sqrt(x * x + y * y);
}

You can use next condition:
x^2+y^2<R^2
Where R - radius,
All this points are in circle.

Drawing Images to fit circle

I'm making an inventory screen for a game I'm working on, and I'd like to be able to draw a series of panels representing each item in the inventory. I want to be able to fit those panels on a circular path.
Here's a mock up I made of what I mean
http://fc02.deviantart.net/fs70/f/2010/280/7/2/khmp_magic_menu_concept_by_magna_alphamon-d30a7em.png
basically I'd like to be able to, give a radius, a center point, and the y co-ordinate to start drawing at, draw this series of panels so they align with the path of the circle like in the image.
Computing the y dimension is easy, its just the startposition y + panel height * panel index, but I'm unsure how to compute the x for a variable radius/center point circle.
Any help would be appreciated.
This is in C#, but something similar in C/C++ will be fine as long as I can convert it
Thanks in advance
EDIT: To calirify, y's position is relative to the top or bottom of the screen and is independent of the circle. If a given y does not map to a point on the circle, then I'll discard that point and not draw the panel.
While ideally I'd like to be able to use any elliptical shape (given two radii), a circle would be good too

Let cx, cy be the coordinates of the center point. Let r be the radius of the circle. Let y be the drawing y-coordinate and x, the x-coordinate. You observe that y = cy + panel height * panel index. By the magic of right triangles, this means that x^2 + y^2 = r^2. Solving for x, we get x = cx + sqrt(r^2 - (y-cy)^2).
EDIT: Converting to code:
#include <math>
float ordinate(float cx, float cy, float r, float y) {
// assumes cx and cy are in the same coordinate system as x and y
// assumes the coordinate origin is in the lower left corner.
return cx + sqrtf(powf(r,2) - powf(y-cy,2));
}

I'm dumb. After seeing Eric's answer, I remembered I can just rearrange and solve the equations of a circle or elipse as necessary.
Thanks

You can use a rotational matrix for this. Here is a simple algorithm that finds the next point {x, y} such that it is rotated theta radians around a circle. You can start with the first item at x=radius and y=radius (wherever really, just a point that you know will contain an item), and then just continue to increment theta as you loop through your items.
Point Rotate(int x, int y, float theta)
int x_p = (x * Math.Cos(theta)) - (y * Math.Sin(theta));
int y_p = (y * Math.Cos(theta)) + (x * Math.Sin(theta));
return new Point(x_p, y_p);
end
On a side note; I always preferred "Bolt1, Bolt2, Bolt3" to "Thunder, Thundara, Thundaga" =P