Making moving eyes in java

Question

i have been trying to make a java applet in which:
there is a circle as an eye (using the drawOval())
and then there is an eyeball (using the fillOval())
the eyeball is supposed to follow the movements of the mouse cursor...
and i m not able think of any idea to do that...
can anyone please give me an idea as to how to do this??

Answer 1

the code i posted earlier was opaque and unclear, so this time i post a clearer code
//moving eyeball
import java.awt.*;
import java.awt.event.*;
import java.applet.*;
public class eyeball extends Applet implements MouseMotionListener
//(x1,y1)=mouse co-ordiantes
//(x,y)=eyeball co-ordianates
//(cx,cy)=eye-centre co-ordinates
//r=radius of eye
int x1,y1,x,y,cx,cy,r;
double m,k1,k2,k3,k4,k5,k6,k7,k8,pvx,pvy,pvx2,pvy2; //these are intermediate values to eventually get the values of (x,y)
                                                    //and are described below as and when required
public void init()
addMouseMotionListener(this);
public void mouseMoved(MouseEvent e)
x1=e.getX();
y1=e.getY();
cx=330;
cy=330;
r=30;
//"slope_numerator" is the numerator part in the formula of slope of the line
//(m=y2-y1/x2-x1) the line here is the line joining the centre of the cirle to
//the co-ordinates of the mouse
double slope_numerator=(y1-cy);
//"slope_dinominator" is the dinominator part in the formula of slope
//(m=y2-y1/x2-x1) the line here is the line joining the centre of the cirle to
//the co-ordinates of the mouse     
double slope_dinominator=(x1-cx);      
//slope of the line
m=slope_numerator/slope_dinominator;   
//start of finding values of x
//from the equation y-y1=m(x-x1) we make y the subject of formula . i.e. y = mx - (mx1-y1)
//so here k1=mx1-y1
//so the equation of line is y=mx-k1
//product_m_and_x1 is the product of m*x1
double product_m_and_x1=m*x1;
//so ,k1 difference between product_m_and_x1 and y1
k1= product_m_and_x1-y1;
//from the equation of circle r^2=(cx-x)^2 +(cy-y)^2 (where r=radius and r^2 => (r squared and so on)
//now i substitute equation of line(y=mx-k1)  in equation of circle
//r^2=(cx-x)^2 + (cy-(mx-k1))^2
//i group (k1+cy) together as k2 ie k2=k1+cy
//so the equation becomes
//r^2=(cx-x)^2+(k2-mx)^2..............................................................................(equation2)
k2=cy+k1;
//when the equation2 is expanded we get
//r^2=[(cx)^2  +  (k2)^2]             -        x(2*cx+2*(k2)m)        +        x^2(1+m^2)...............................(equation3)
//taking k3=[(cx)^2+(k2)^2] ie.sum of cx_squared and k2_squared
//cx_squared is square of cx value
double cx_squared=cx*cx;
//k2_squared is square of k2 value
double k2_squared=k2*k2;
k3=cx_squared+k2_squared;
//from equation3(r^2=(cx)^2+(k2)^2-x(2*cx+2*(k2)m)+x^2(1+m^2))
//take k4=[2*(cx)+2*(k2)*m] ie.sum of 2 times cx and 2*m times k2
//cx_times_2 is 2*cx
double cx_times_2=2*cx;
//k2_times_2_times_m is 2*k2*m
double k2_times_2_times_m=2*k2*m;
//therefore calculating k4
k4=cx_times_2 + k2_times_2_times_m;
//from equation3(r^2=(cx)^2+(k2)^2-x(2*cx+2*(k2)m)+x^2(1+m^2))
//take k5=[1+m^2] ie sum of m_squared and 1
//first calculate m_squared as m*m
double m_squared=m*m;
//calculate k5=[1+m^2]
k5=1+m2_squared;
//so the equation3 is reduced to r^2=k3 - x(k4) + x^2(k5)
//bringing r^2 to Right Hand Side(RHS) ....
//0=[k3-r^2]    -     x(k4)      +      x^2(k5)...............................................................................(equation4)
//k6=[k3-r^2]
//Therefore the quadratic equation in x becomes
//x^2(k5) - x(k4) +k6 = 0.....................................................................................(equation5)
//first we calculated k6
//for that we need value of r_squared
//therefore we calculate r_squared = r*r
double r_squared=r*r;
//then we find k6=k3-r^2 in equation4
k6=k3-r2;
//we now solve the quadratic equation in x of the (equation5)
//for that we use the standard formula used to get the value of x in the equation
// (a)x^2+(b)x+c=0
//which is x= {-b (-+) squareroot of (b_squared - 4*a*c)}/(whole divided by 2*a)
//in (equation5):: a=k5, b=-k4 , c=k6
//first we find the value of the determinant ie. squareroot of (b_squared - 4*a*c)
//so since k4 is b here we find the value of k4_squared
double k4_squared=k4*k4;
//then we find the value of 4*a*c
//so since a=k5 and c=k6 we store the value of product into the variable k5k6
double k5k6=4*k5*k6;                 //(ok u guys were right the coding earlier was horrible. there was a mistake here)
//now we find the difference of b_squared and 4*a*c ie. k4_squared and k5k6
double discriminant_squared=k42-k5k6;
//then we find the discriminant_value by taking square root of discriminant_squared
discriminant_value=Math.sqrt(discriminant_squared);
//we then find the dinominator_value in the formula. ie. 2*a
//the value of a here is k4. so dinominator_value becomes 2*k4
dinominator_value=2*k4;
//then we calculate the value of => (-b - discriminant_value )
//here that would be difference between k4 and discriminant_value
double numerator_value1=k4-discriminant_value;
//then we find the first possible value of x (pvx)
//the first possible value of x(pvx) is numerator_value1/dinominator_value
//since, numerator_value1 represents [-b - squareroot of (b_squared - 4*a*c)]
pvx=numerator_value1/dinominator_value;
//then we calculate the value of => (-b + discriminant_value )
//here that would be sum of k4 and discriminant_value
double numerator_value2=k4+k7;
////then we find the second possible value of x (pvx2)
//the second possible value of x(pvx2) is numerator_value2/dinominator_value
//since, numerator_value2 represents [-b + squareroot of (b_squared - 4*a*c)]
pvx2=numerator_value2/dinominator_value;
//So the two possible x values are pvx,pvx2
//end of finding values of x
//start of finding values of y
//from the equation y-y1=m(x-x1) we made y the subject of formula . i.e. y = mx - (mx1-y1) or y=mx -k1
//now we have the value of x as pvx and pvx2 so now we can find values of y
//product_mx1 represents the product of mx in the equation y=mx-k1 when x is pvx
double product_mx1=m*pvx;
//product_mx2 represents the product of mx in the equation y=mx-k1 when x is pvx2
double product_mx2=m*pvx2;
//so now we can calculate the possible value of y(pvy and pvy2) corresponding to possible value of x(pvx and pvx2)
//so pvy is product_mx1 - k1 form the equation y=mx - k1
pvy=product_mx1-k1;
//similarly pvy2 is product_mx2 -k1 from the equation of line y=mx-k1
pvy2=product_mx2-k1;
//end of finding values of y
//but the values of (pvx,pvy) and (pvx2,pvy2) are in double , so convert them to int
//therefore
x=(int) pvx;
y=(int) pvy;
//so the co-ordinates of the eyeball should be (x,y)
//(x,y) is the point on the eye(circle) which satisfies the line joining the mouse pointer co-ordinates and centre of eye(circle)
//end of calculations
repaint();
//end of mouseMoved method
public void mouseDragged(MouseEvent e){}
public void paint(Graphics g)
//the eye
g.drawOval(300,300,60,60); 
//the eyeball
g.fillOval(x,y,20,20);
//the line joining the centre of eye(circle) to the position of mousepointer
g.drawLine(330,330,x1,y1);
//calculated values for the sake of reference
g.drawString("x1,y1="+x1+"::::"+y1,50,50);                  //this shows current co-ordinates of mouse pointer
g.drawString("m="+m,50,65);                                 //this shows the slope(m) of the line y=mx-k1
g.drawString("k1="+k1,50,80);                               //this shows the value of k1 
g.drawString("k2="+k2,50,95);                               //this shows the value of k2           
g.drawString("k3="+k3,50,110);                              //this shows the value of k3
g.drawString("k4="+k4,50,125);                              //this shows the value of k4
g.drawString("k5="+k5,50,140);                              //this shows the value of k5
g.drawString("k6="+k6,50,155);                              //this shows the value of k6
g.drawString("k7="+k7,50,170);                              //this shows the value of k7
g.drawString("k8="+k8,50,185);                              //this shows the value of k8
g.drawString("pvx,pvy="+pvx+"::::"+pvy,50,200);             //this shows current co-ordinates (pvx,pvy)
g.drawString("pvx2,pvy2="+pvx2+"::::"+pvy2,50,220);         //this shows current co-ordinates (pvx2,pvy2)
g.drawString("x,y="+x+"::::"+y,50,240);                     //this shows current co-ordinates of the eye-ball(fillOval)
//end of paint method
//end of applet
/*<applet code="eyeball" height=800 width=600></applet>*/