Shifting Circles

The left edge of the leftmost circle is 0 pixels from the left edge. The right edge of the rightmost circle is 30 - diameter. pixels from the edge. If every circle was shifted over half this distance, the drawing would be fixed. This can be done, by adding yet more arithmetic to the applet:

import javax.swing.JApplet;
import java.awt.*;

// Assume that the drawing area is 300 by 150.
// Draw ten red circles side-by-side across the drawing area.
public class TenShiftedCircles extends JApplet
{
  final int width = 300, height = 150;

  public void paint ( Graphics gr )
  { 
    gr.setColor( Color.red );
    int diameter = 20;
    int Y = height/2 - diameter/2;      // the top edge of the squares
    int shift = (width/10-diameter)/2;  // shift circles right

    int count =  0 ;
    while (  count < 10  )
    {
      int X = count*width/10+shift;          // the left edge of each square
      gr.drawOval( X, Y, diameter, diameter );
      count = count + 1; 
    }
  }
}

Here is what it outputs:

The arithmetic used in drawing these circles is getting somewhat messy. It is not, for example, very clear what the statement

int shift = (width/10-diameter)/2;  // shift circles right

does, even after it has been explained.