Let's say that I have a win form app and I have a picture box called pictureBox1. Then I run the following code:
public System.Drawing.Graphics graphics;
public System.Drawing.Pen blackPen = new System.Drawing.Pen(Color.Black, 2);
public void drawVennDiagram()
{
graphics = pictureBox1.CreateGraphics();
graphics.DrawEllipse(blackPen, 0, 0, 100, 100);
graphics.DrawEllipse(blackPen, 55, 0, 100, 100);
}
If I call drawVennDiagram() , it will draw two circles in pictureBox1 and the circles overlap just enough to look like a venn diagram.
What I am trying to achieve is as follows:
Run Method
Aif the mouse clicks anywhere outside of the venn diagram but in the picturebox.Run Method
Bif the mouse clicks only inside of the first circle.Run Method
Cif the mouse clicks inside of both circles.Run Method
Dif the mouse clicks only inside of the second circle.
So far I have written the code below, which essentially tracks where the cursor clicks, but I have no way of figuring out which parameter (a, b, c, d) the cursor location follows.
private void pictureBox1_Click(object sender, EventArgs e)
{
this.Cursor = new Cursor(Cursor.Current.Handle);
int xCoordinate = Cursor.Position.X;
int yCoordinate = Cursor.Position.Y;
}
private void pictureBox1_MouseUp(object sender, MouseEventArgs e)
{
int xCoordinate = e.X;
int yCoordinate = e.Y;
}
What is the best way to achieve this?
Yes. Given a circle is positioned with center (xc,yc) and radius r, a coordinate (x,y) is within the circle if: (x-xc)2+(y-yc)2≤r2.
Given we know that, we also know that your circles have a center at (50,50) and (105,50) and each have a radius of 50. So now we define a method:
public static bool InsideCircle (int xc, int yc, int r, int x, int y) {
int dx = xc-x;
int dy = yc-y;
return dx*dx+dy*dy <= r*r;
}
Now you can use:
private void pictureBox1_MouseUp(object sender, MouseEventArgs e) {
int x = e.X;
int y = e.Y;
bool inA = InsideCircle(50,50,50,x,y);
bool inB = InsideCircle(105,50,50,x,y);
if(inA && inB) {
C();
} else if(inA) {
B();
} else if(inB) {
D();
} else {
A();
}
}
Note however that for now, the two circles you paint do not overlap anyway.
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