Get the leftmost point in a rotated GraphicsPath

Question

I create a GraphicsPath object, add a ellipse, rotate the GraphicsPath object and then draw it. Now I want to get for example the leftmost Point of the graphicsPath has so I can check if it is within certain boundaries (users can move the graphicsPath with the mouse).

I am currently using the GetBounds() method from the GraphicsPath but that only result in the following

.

The blue is the rectangle from GetBounds() so as you can tell there is some space between the leftmost Point i get from this method compared to the Point I want. How can I get the Point I'm looking for instead?

Answer 1

If you actually rotate the GraphicsPath you can use the Flatten function to aquire a large number of path points. Then you can select the minimum x-value and from it the corresponding y-value.

This will work since you have an ellipse, so only one point can be the most leftmost..

private void panel1_Paint(object sender, PaintEventArgs e)
{
    GraphicsPath gp = new GraphicsPath();
    gp.AddEllipse(77, 55, 222, 77);

    Rectangle r = Rectangle.Round(gp.GetBounds());
    e.Graphics.DrawRectangle(Pens.LightPink, r);
    e.Graphics.DrawPath(Pens.CadetBlue, gp);

    Matrix m = new Matrix();
    m.Rotate(25);
    gp.Transform(m);
    e.Graphics.DrawPath(Pens.DarkSeaGreen, gp);
    Rectangle rr = Rectangle.Round(gp.GetBounds());
    e.Graphics.DrawRectangle(Pens.Fuchsia, rr);

    GraphicsPath gpf = (GraphicsPath)gp.Clone();
    gpf.Flatten();
    float mix = gpf.PathPoints.Select(x => x.X).Min();
    float miy = gpf.PathPoints.Where(x => x.X == mix).Select(x => x.Y).First();
    e.Graphics.DrawEllipse(Pens.Red, mix - 2, miy - 2, 4, 4);
}

Please don't ask me why the rotated bounds are so wide - I really don't know!

If instead you rotate the Graphics object before drawing you may still use the same trick..