Category: Solutions

Below is the solution for Algorithm Problem 1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

#define PI 3.14159265358979323846

float compute_comman_area(double r0,double r1,double x0,double y0,double x1,double y1){

     double c,CBA,CAB,CBD,CAD,Area;

     //length
     c = sqrt(pow((x1-x0),2) + pow((y1-y0),2));

     //IN RADIANS
     CBA = (acos((pow(r1,2) + pow(c,2) - pow(r0,2))/(2*r1*c)));
     CAB = (acos((pow(r0,2) + pow(c,2) - pow(r1,2))/(2*r0*c)));

     CBD = 2*CBA;
     CAD = 2*CAB;

     Area = (CBD*pow(r1,2) - pow(r1,2)*sin(CBD) + CAD*pow(r0,2) - pow(r0,2)*sin(CAD))/2;

     printf("%lf", CBA);

     return Area;
}

int main(){

    double r1=10,r0=10,x0=0,y0=0,x1=0,y1=10, Area, Area0;

    //getting input
    /*printf("Enter the radius of the first circle\nand the centre point");
    scanf("%lf%lf%lf",&r0,&x0,&y0);
    printf("Enter the radius of the first circle\nand the centre point");
    scanf("%lf%lf%lf",&r1,&x1,&y1);*/

    //calling function to compute area
    Area0 = PI * pow(r0, 2) + PI * pow(r1, 2);
    Area = Area0 - compute_comman_area(r0,r1,x0,y0,x1,y1);

    //printing the area
    printf("The area thus computed is:%lf",Area);

return 0;
}

You can download the source file here.

If you will go for the problem statement then you will find that you have to find the Area covered by intersection of two or three circles and the coordinates and radius is given.

Lets see how to find the shared area between intersection of two circles.

Let two circles are given.

1. center(x0,y0), radius (r0)
 2. center(x1,y1), radius (r1)

Let A be the centre of the circle ( x0, y0 ) and B be the centre of the other circle ( x1,y1 ).

Draw the circles with appropriate radii r0 and r1 so that there is a reasonable amount of overlap.  The length AB is calculated from the coordinates of the centre:

AB = sqrt{(x1-x0)^2 + (y1-y0)^2}

For convenience let this length be denoted by c.

The two circles intersect in two points which I will label C and D.

Now we must calculate the angles CAD and CBD, and we do this using the cosine formula. In fact it is half of these angles that we first calculate, using triangle CAB.

r0^2 = r1^2 + c^2 - 2*r1*c*cos(CBA)

cos(CBA) = (r1^2 + c^2 - r0^2)/(2*r1*c)

found CBA, then CBD = 2(CBA).

Similarly,
cos(CAB) = (r0^2 + c^2 - r1^2)/(2*r0*c)
and then
CAD = 2(CAB)

Express CBD and CAD in radian measure. Then we find the segment of each of the circles cut off by the chord CD, by taking the area of the sector of the circle BCD and subtracting the area of triangle BCD.
Similarly we find the area of the sector ACD and subtract the area of triangle ACD.

Area = ( 1/2 )( CBD ) r1 ^ 2 - ( 1/2 ) r1 ^ 2 * sin( CBD )
  + ( 1/2 )( CAD ) r0 ^ 2 - ( 1/2 ) r0 ^ 2 * sin( CAD )

Remember that for the area of the sectors you must have CBD and CAD in radians.

One more thing if the two circles are of the SAME radius please note that the area is symmetrical about the chord CD. Therefore, you only need to find the area in one half of the intersection and multiply by 2.

A shorter equation is

Area = 2 * ( ( 1/2 ) ( CBD ) r1 ^ 2 - ( 1/2 ) r1 ^ 2.sin( CBD ) ).

One more derivation if the redis is equal then you can find out using, this formula
Area = r^2*(q - sin(q))  where q = 2*acos(c/2r),
where c = distance between centers and r is the common radius.

I think now you can solve this. If something goes wrong, then post in the comments.