学步园

There is a cycle with its center on the origin.
Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other
you may assume that the radius of the cycle will not exceed 1000.

There are T test cases, in each case there are 2 decimal number representing the coordinate of the given point.

For each testcase you are supposed to output the coordinates of both of the unknow points by 3 decimal places of precision

Alway output the lower one first(with a smaller Y-coordinate value), if they have the same Y value output the one with a smaller X.

2
1.500 2.000
563.585 1.251

0.982 -2.299 -2.482 0.299
-280.709 -488.704 -282.876 487.453
#include<stdio.h>
#include<math.h>
int main()
{
    int t;
    double x[3],y[3],c[2],a,b;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf",&x[0],&y[0]);
        c[0]=x[0]*x[0]+y[0]*y[0];
        if(x[0]==0)
        {
            y[1]=y[2]=-y[0]/2;
            x[1]=-sqrt(c[0]-y[1]*y[1]);
            x[2]=sqrt(c[0]-y[2]*y[2]);
        }
        else
        {
            c[1]=-c[0]+c[0]*c[0]/(4*x[0]*x[0]);
            b=c[0]*y[0]/(x[0]*x[0]);
            a=1+y[0]*y[0]/(x[0]*x[0]);
            y[1]=(-b-sqrt(b*b-4*a*c[1]))/(2*a);
            y[2]=(-b+sqrt(b*b-4*a*c[1]))/(2*a);
            x[1]=-c[0]/(2*x[0])-y[0]*y[1]/x[0];
            x[2]=-c[0]/(2*x[0])-y[0]*y[2]/x[0];
        }
        printf("%.3lf %.3lf %.3lf %.3lf\n",x[1],y[1],x[2],y[2]);
    }
}