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Explanation on libgdx draw method

I am learning libgdx, I have passed around some of tutorials but I always face problems understanding the batch.draw() method
public void draw(Texture texture,
float x,
float y,
float originX,
float originY,
float width,
float height,
float scaleX,
float scaleY,
float rotation,
int srcX,
int srcY,
int srcWidth,
int srcHeight,
boolean flipX,
boolean flipY)
I read its documentation and still got confused of the following statement:-
The rectangle is offset by originX, originY relative to the origin.
WHAT DOES THIS MEAN?. Which origin are talking about here?
Also i did a simple sketch of what I visual understand the draw() method. Am I on the right path?
Thanks.

The rectangle is offset by originX, originY relative to the origin.
Scaling and Rotation performed around originX, originY.
batch.begin();
batch.draw(texture,300,200,50,50,100,100,1,1,0,100,100,50,50,false,false);
batch.end();
And Output is
Now rotate by 90 degree :
batch.draw(texture,300,200,50,50,100,100,1,1,90,100,100,50,50,false,false);
so rectangle offseted around centre by 90 degree
originX and originY is in the center of Rectangle so no offset appears in x,y
Let's put originX and originY at left bottom of rectangle and rotate by 90 degree
batch.draw(texture,300,200,0,0,100,100,1,1,90,100,100,50,50,false,false);

java slick2d image rotation according to current mouse cursor location

float mouseX = input.getMouseX();
float mouseY = input.getMouseY();
float xDistance = mouseX - imageX;
float yDistance = mouseY - imageY;
double angleToTurn = Math.toDegrees(Math.atan2(yDistance, xDistance));
image.setRotation((float) angleToTurn);
the image is rotating but the mouse's cursor is off the center of the image, it's always on the right side or x coordinate and y coordinates of the image. When I drag the mouse 90 degrees to the right, the image will rotate 90 degrees to the right but the mouse's cursor is out of the center of the image, it's always on the right hand side of the image wherever the rotation occur. Should I make an if statement? Thank you in advance.

How to align rectangle to bottom of any device?

rec.set(0, 0, canvas.getWidth(), canvas.getHeight()/2);
It creates the rectangle as expected, I know 0, 0 is x, y but how do I set y to the bottom so it stays at the bottom on all devices with different size?

you haven't given much information, so I'll state my assumptions
the x,y would be the top-left of your shape, so assuming that the canvas is the full height of the device:
x - 0, as you want the shape to start on the left edge, y - the height of the device minus the height of the shape
your code would be
int width = canvas.getWidth();
int height = canvas.getHeight()/2;
int x = 0;
int y = canvas.getHeight() - height;
rec.set(x, y, width, height);

Work out the maximum size of an image?

I am creating a little game in Java and I have an image which gets rotated.
As you can see in the two images below, there is a giant ship which slowly rotates in the game, but when it gets to a certain point it gets cut off (due to its own little BufferedImage).
Heres my rendering code:
public void drawImageRotated(BufferedImage img, double x, double y, double scale, double angle) {
x -= xScroll;
y -= yScroll;
BufferedImage image = new BufferedImage((int)(img.getWidth() * 1.5D), (int)(img.getHeight() * 1.5D), 2);
Graphics2D g = (Graphics2D)image.getGraphics();
g.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON);
g.rotate(Math.toRadians(angle), image.getWidth() / 2, image.getHeight() / 2);
g.drawImage(img, image.getWidth() / 2 - img.getWidth() / 2, image.getHeight() / 2 - image.getHeight() / 2, null);
g2d.drawImage(image, (int)(x-image.getWidth()*scale/2), (int)(y-image.getHeight()*scale/2), (int)(image.getWidth()*scale), (int)(image.getHeight()*scale), null);
g.dispose();
}
Back to the matter at hand, how can i work out the maximum x and y size of an image during rotation so I can compensate with my buffered images size?

If you have a basically rectangular image which is rotated around its center, the maximum width and height during rotation will be when a diagonal of the image rectangle is horizontal or vertical. This diagonal distance could be computed with the Pythagorean Theorem and used for the width and height of the BufferedImage.
int size = (int) Math.sqrt((img.getWidth() * img.getWidth()) + (img.getHeight() * img.getHeight()));
BufferedImage image = new BufferedImage(size, size, 2);
// The rest of your code as before

how can i work out the maximum x and y size of an image during rotation so I can compensate with my buffered images size?
double sin = Math.abs(Math.sin(angle));
double cos = Math.abs(Math.cos(angle));
int w = image.getWidth();
int h = image.getHeight();
int neww = (int)Math.floor(w*cos+h*sin);
int newh = (int)Math.floor(h*cos+w*sin);
The above code was taken from this example: Java(SWING) working with Rotation

An alternative is to rotate the actual Graphics object, draw the image, and restore the rotation:
AffineTransform old = g2d.getTransform();
g2d.rotate(Math.toRadians(angle), x + image.getWidth() / 2, y + image.getWidth() / 2);
g2d.drawImage(image, x, y, null);
g2d.setTransform(old);

Let's consider width being the width of the original image, height its original height and angle the rotation angle value in radians.
According to my calculations, the size of the rotated image is something like this:
rotatedWidth = Math.cos(angle) * width + Math.sin(angle) * height;
rotatedHeight = Math.sin(angle) * width + Math.cos(angle) * height;
You may also need to take a look at this thread as well, as it may help.

How to approximate an ellipse to fill a given rectangle, using bezier curves?

I tried the code below, which draws a good approximation of a circle if the rectangle's width is the same as its height; but it doesn't draw a great oval, the "corners" are very pointed. Any suggestions?
float width = rect.width();
float height = rect.height();
float centerX = rect.width() / 2;
float centerY = rect.height() / 2;
float diameter = Math.min(width, height);
float length = (float) (0.5522847498 * diameter/2);
path.moveTo(0, centerY);
path.cubicTo(0, centerY - length, 0, centerX - length, 0, centerX, 0);
path.cubicTo(centerX + length, 0, width, centerY - length, height, centerY);
path.cubicTo(width, centerY + length, centerX + length, height, centerX, height);
path.cubicTo(centerX - length, height, 0, centerY + length, 0, centerY);

You should scale length according to which axis it's on, so that the distance from each arc endpoint to the adjacent control points is (not fixed but) a fixed fraction of the axis you're moving parallel to at that point.

If it's a true rectangle, with right angles, then it should be easy. The major and minor axes of the ellipse equal the lengths of the sides of the rectangle, and the center of the ellipse is located at the intersection of the rectangle's diagonals.
I don't know how to express it as Bezier splines off the top of my head, but the classic equation for an ellipse should be easy enough to write, as long as you transform to the appropriate coordinate system first (e.g. x-axis along the major axis of the rectangle/ellipse, y-axis along the minor axis of the rectangle/ellipse).