Too hard
Posted: Mon Mar 28, 2005 8:10 pm
I think you've done it too complicatedly. Try to solve the eqations for thee points - you will need only basic mathemathic operations +-* and/. That'd be more precise.
Page 4 of 9
Posted: Tue Mar 29, 2005 12:52 am
Also, hypot returns an "int" in ASCII C unless you explicitly declare the function header.
190 - Circle Through Three Points
Posted: Thu Apr 28, 2005 3:14 pm
what is wrong with my solution
it gets wrong answer
my code is very simple
so anybody can look at a glance
have any idea why wa
or have any input for which my solution produce wrong
here is my code.
thanks
it gets wrong answer
my code is very simple
so anybody can look at a glance
have any idea why wa
or have any input for which my solution produce wrong
here is my code.
thanks
Code: Select all
#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
int main (void)
{
float x1, x2, x3, y1, y2, y3;
float c, d, e, h, k, r;
while(scanf("%f%f%f%f%f%f",&x1,&y1,&x2,&y2,&x3,&y3)==6){
c=((x2*x2+y2*y2-x3*x3-y3*y3)*(y1-y2) - (x1*x1 + y1*y1 - x2*x2 - y2*y2)*(y2-y3))/((x1-x2)*(y2-y3)-(x2-x3)*(y1-y2));
d=(x1*x1+y1*y1-x3*x3-y3*y3-c*(x3-x1))/(y3-y1);
e=-1*(x1*x1+y1*y1+c*x1+d*y1);
h=c/(-2.0);
k=d/(-2.0);
r=sqrt(h*h+k*k-e);
printf("(x %c %.3f)^2 + (y %c %.3f)^2 = %.3f^2\n",(h<0)?'+':'-', (h<0)?(-h):h, (k<0)?'+':'-',(k<0)?(-k):k,r);
printf("x^2 + y^2 %c %.3fx %c %.3fy %c %.3f = 0\n\n",(c<0)?'-':'+',(c<0)?(-c):c, (d<0)?'-':'+',(d<0)?(-d):d, (e<=0)?'-':'+', (e<0)?(-e):e);
}
return 0;
}
Posted: Fri Apr 29, 2005 11:37 pm
Hi emotional blind !! try with this input
in:
out:
Hope it helps
Keep posting !!
in:
Code: Select all
5 5 4 7 16 5
Code: Select all
(x - 10.500)^2 + (y - 9.000)^2 = 6.801^2
x^2 + y^2 - 21.000x - 18.000y + 145.000 = 0
Keep posting !!
Posted: Mon May 02, 2005 11:32 am
dear Ghust_omega
thanks a lot ...................................
I got ac
thanks again
thanks a lot ...................................
I got ac
thanks again
Posted: Mon May 02, 2005 11:33 am
dear Ghust_omega
thanks a lot ...................................
I got ac
thanks again
thanks a lot ...................................
I got ac
thanks again
problem 190
Posted: Tue May 24, 2005 11:02 pm
what is wrong with my solution
it gets wrong answer
my code is very simple
so anybody can look at a glance
have any idea why wa
or have any input for which my solution produce wrong
here is my code.
thanks
#include<stdio.h>
#include<math.h>
int z;
float x[3],y[3];
double ak,bk,ck;
void output(void)
{
double r;
r=sqrt(ak*ak+bk*bk-ck);
printf("(x ");
if (-1*ak>0) printf("- ");
else printf("+ ");
printf("%.3f)^2 + (y ",fabs(ak));
if (-1*bk>0) printf("- ");
else printf("+ ");
printf("%.3f)^2 = %.3f",fabs(bk),r);
if (r*r!=r) printf("^2\n");
else printf("\n");
printf("x^2 + y^2 ");
if (2*ak<0) printf("- ");
else printf("+ ");
printf("%.3fx ",fabs(2*ak));
if (2*bk<0) printf("- ");
else printf("+ ");
printf("%.3fy ",fabs(2*bk));
if (ck<0) printf("- ");
else printf("+ ");
printf("%.3f = 0",fabs(ck));
}
void hitung()
{
int i;
double a[6],b[6],p[2],bil[6];
for (i=0;i<3;i++) {
bil=x*x+y*y;
a=x*2;
b=y*2;
}
for (i=3;i<5;i++) {
bil=bil[i-3]-bil[i-2];
a[i]=a[i-3]-a[i-2];
b[i]=b[i-3]-b[i-2];
}
p[0]=a[3];
p[1]=a[4];
for (i=3;i<5;i++) {
bil[i]=bil[i]*fabs(p[4-i]);
b[i]=b[i]*fabs(p[4-i]);
a[i]=a[i]*fabs(p[4-i]);
}
if((p[0]<0&&p[1]<0)||(p[0]>0&&p[1]>0)) {
bil[5]=bil[3]-bil[4];
b[5]=b[3]-b[4];
} else {
bil[5]=bil[3]+bil[4];
b[5]=b[3]+b[4];
}
if(b[5] == 0)
{
bk=0.0;
} else{
bk=-1*bil[5]/b[5];}
if(a[4]== 0.0){
ak=0.0;} else{
ak=(bil[4]*-1+bk*b[4]*-1)/a[4];
}
ck=bil[0]*-1+ak*a[0]*-1+bk*b[0]*-1;
}
main()
{
while (!feof(stdin)) {
fscanf(stdin,"%f %f %f %f %f %f",&x[0],&y[0],&x[1],&y[1],&x[2],&y[2]);
hitung();
output();
printf("\n\n");
}
return 0;
}
it gets wrong answer
my code is very simple
so anybody can look at a glance
have any idea why wa
or have any input for which my solution produce wrong
here is my code.
thanks
#include<stdio.h>
#include<math.h>
int z;
float x[3],y[3];
double ak,bk,ck;
void output(void)
{
double r;
r=sqrt(ak*ak+bk*bk-ck);
printf("(x ");
if (-1*ak>0) printf("- ");
else printf("+ ");
printf("%.3f)^2 + (y ",fabs(ak));
if (-1*bk>0) printf("- ");
else printf("+ ");
printf("%.3f)^2 = %.3f",fabs(bk),r);
if (r*r!=r) printf("^2\n");
else printf("\n");
printf("x^2 + y^2 ");
if (2*ak<0) printf("- ");
else printf("+ ");
printf("%.3fx ",fabs(2*ak));
if (2*bk<0) printf("- ");
else printf("+ ");
printf("%.3fy ",fabs(2*bk));
if (ck<0) printf("- ");
else printf("+ ");
printf("%.3f = 0",fabs(ck));
}
void hitung()
{
int i;
double a[6],b[6],p[2],bil[6];
for (i=0;i<3;i++) {
bil=x*x+y*y;
a=x*2;
b=y*2;
}
for (i=3;i<5;i++) {
bil=bil[i-3]-bil[i-2];
a[i]=a[i-3]-a[i-2];
b[i]=b[i-3]-b[i-2];
}
p[0]=a[3];
p[1]=a[4];
for (i=3;i<5;i++) {
bil[i]=bil[i]*fabs(p[4-i]);
b[i]=b[i]*fabs(p[4-i]);
a[i]=a[i]*fabs(p[4-i]);
}
if((p[0]<0&&p[1]<0)||(p[0]>0&&p[1]>0)) {
bil[5]=bil[3]-bil[4];
b[5]=b[3]-b[4];
} else {
bil[5]=bil[3]+bil[4];
b[5]=b[3]+b[4];
}
if(b[5] == 0)
{
bk=0.0;
} else{
bk=-1*bil[5]/b[5];}
if(a[4]== 0.0){
ak=0.0;} else{
ak=(bil[4]*-1+bk*b[4]*-1)/a[4];
}
ck=bil[0]*-1+ak*a[0]*-1+bk*b[0]*-1;
}
main()
{
while (!feof(stdin)) {
fscanf(stdin,"%f %f %f %f %f %f",&x[0],&y[0],&x[1],&y[1],&x[2],&y[2]);
hitung();
output();
printf("\n\n");
}
return 0;
}
Posted: Wed Jun 08, 2005 12:54 pm
Hi, Oyhama Hora,
First, suppose we have found the equation of the circle
in the following form:
Here we have that R>0 but A and B could have any sign
( they can be zeros as well ).
Then let's suppose that your algorithm is OK so that you
find the correct values for A and B.
Then my question is: what do you print if A==0 ?
And same question for B: if B==0 what do you print ?
Looking at your code ( I haven't compiled and run it )
I think you will print something like:
which is not correct. Correct output should be
That means the judge wants NO brackets and NO 0.000
in the case A==0 ( B==0 ).
See also the two test cases I have posted here:
http://online-judge.uva.es/board/viewtopic.php?p=35401
I hope this makes the things clear.
Try to change this and see if you still get WA.
First, suppose we have found the equation of the circle
in the following form:
Code: Select all
(X + A)^2 + (Y + B)^2 = R^2( they can be zeros as well ).
Then let's suppose that your algorithm is OK so that you
find the correct values for A and B.
Then my question is: what do you print if A==0 ?
And same question for B: if B==0 what do you print ?
Looking at your code ( I haven't compiled and run it )
I think you will print something like:
Code: Select all
(X + 0.000)^2 ... Code: Select all
X^2 ... in the case A==0 ( B==0 ).
See also the two test cases I have posted here:
http://online-judge.uva.es/board/viewtopic.php?p=35401
I hope this makes the things clear.
Try to change this and see if you still get WA.
Posted: Wed Jun 08, 2005 1:09 pm
I think the two test cases which follow may be helpful
to anyone who still has WA on this problem.
INPUT
OUTPUT
Good luck to everyone !
to anyone who still has WA on this problem.
INPUT
Code: Select all
7 0 11 0 9 2
0 9 0 15 3 12
Code: Select all
(x - 9.000)^2 + y^2 = 2.000^2
x^2 + y^2 - 18.000x + 77.000 = 0
x^2 + (y - 12.000)^2 = 3.000^2
x^2 + y^2 - 24.000y + 135.000 = 0
why i get worng answer?
Posted: Sun Jul 31, 2005 11:35 am
why i get wrong answer or compile error
this is my code:
#include<stdio.h>
#include<math.h>
int main()
{
double x1,x2,x3,y1,y2,y3,h,k,r,d1,e;
while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)==6)
{
d1=2*((x2-x1)*(y3-y2)-(x3-x2)*(y2-y1));
h=((y2-y1)*(x3*x3+y3*y3-x2*x2-y2*y2)-(y3-y2)*(x2*x2+y2*y2-x1*x1-y1*y1))/d1;
k=((x3-x2)*(x2*x2+y2*y2-x1*x1-y1*y1)-(x2-x1)*(x3*x3+y3*y3-x2*x2-y2*y2))/d1;
r=sqrt(pow((x1-(-1.0*h)),2)+pow((y1-(-1.0*k)),2));
e=r*r-(h*h+k*k);
if( fabs(h) < 1e-8 && fabs(k)<1e-8)
{
printf("(x + 0.000)^2 + (y + 0.000)^2 = 0.000^2\n",r);
printf("x^2 + y^2 + 0.000x + 0.000y + 0.000 = 0\n",e);
}
else if( h <0 && k <0 && e < 0)
{
h=-1.0*h;
k=-1.0*k;
e=-1.0*e;
printf("(x - %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx - %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h < 0 && k< 0 && e>=0)
{
h=-1.0*h;
k=-1.0*k;
printf("(x - %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx - %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if( h >=0 && k <0 && e<0)
{
k=-1.0*k;
e=-1.0*e;
printf("(x + %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx - %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h>=0 && k<0 && e>=0)
{
k=-1.0*k;
printf("(x + %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx - %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h <0 && k>=0 && e< 0)
{
e=-1.0*e;
h=-1.0*h;
printf("(x - %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx + %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h<0 && k>=0 && e>=0)
{
h=-1.0*h;
printf("(x - %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx + %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h>=0 && k>=0 && e<0)
{
printf("(x + %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx + %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else//(h>=0 && k>=0 && e>=0)
{
printf("(x + %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx + %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
printf("\n");
}
return 0;
}
this is my code:
#include<stdio.h>
#include<math.h>
int main()
{
double x1,x2,x3,y1,y2,y3,h,k,r,d1,e;
while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)==6)
{
d1=2*((x2-x1)*(y3-y2)-(x3-x2)*(y2-y1));
h=((y2-y1)*(x3*x3+y3*y3-x2*x2-y2*y2)-(y3-y2)*(x2*x2+y2*y2-x1*x1-y1*y1))/d1;
k=((x3-x2)*(x2*x2+y2*y2-x1*x1-y1*y1)-(x2-x1)*(x3*x3+y3*y3-x2*x2-y2*y2))/d1;
r=sqrt(pow((x1-(-1.0*h)),2)+pow((y1-(-1.0*k)),2));
e=r*r-(h*h+k*k);
if( fabs(h) < 1e-8 && fabs(k)<1e-8)
{
printf("(x + 0.000)^2 + (y + 0.000)^2 = 0.000^2\n",r);
printf("x^2 + y^2 + 0.000x + 0.000y + 0.000 = 0\n",e);
}
else if( h <0 && k <0 && e < 0)
{
h=-1.0*h;
k=-1.0*k;
e=-1.0*e;
printf("(x - %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx - %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h < 0 && k< 0 && e>=0)
{
h=-1.0*h;
k=-1.0*k;
printf("(x - %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx - %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if( h >=0 && k <0 && e<0)
{
k=-1.0*k;
e=-1.0*e;
printf("(x + %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx - %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h>=0 && k<0 && e>=0)
{
k=-1.0*k;
printf("(x + %.3lf)^2 + (y - %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx - %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h <0 && k>=0 && e< 0)
{
e=-1.0*e;
h=-1.0*h;
printf("(x - %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx + %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h<0 && k>=0 && e>=0)
{
h=-1.0*h;
printf("(x - %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 - %.3lfx + %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
else if(h>=0 && k>=0 && e<0)
{
printf("(x + %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx + %.3lfy + %.3lf = 0\n",2.0*h,2.0*k,e);
}
else//(h>=0 && k>=0 && e>=0)
{
printf("(x + %.3lf)^2 + (y + %.3lf)^2 = %.3lf^2\n",h,k,r);
printf("x^2 + y^2 + %.3lfx + %.3lfy - %.3lf = 0\n",2.0*h,2.0*k,e);
}
printf("\n");
}
return 0;
}
Posted: Mon Aug 01, 2005 3:20 am
I have three points to make:
1) Which problem is this code supposed to solve?
2) What is the algorithm used?
3) I believe you are solving problem 190. You can search for past threads on problem 190.
1) Which problem is this code supposed to solve?
2) What is the algorithm used?
3) I believe you are solving problem 190. You can search for past threads on problem 190.
Re: still confuse
Posted: Thu Sep 08, 2005 9:43 pm
Dear gracia,gracia wrote:well.. thank you for your answer
and i have tried your advice
but i still got WA
this is my program and may be you can tell me why i got WA
...
int main()
{
while (!feof(stdin)) {
fscanf(stdin,"%f %f %f %f %f %f",&x[0],&y[0],&x[1],&y[1],&x[2],&y[2]);
hitung();
output();
printf("\n\n");
}
return 0;
}
I know it's a little late.
but I think your program has some problem when it reads six numbers from the system.
Because the lack of practice in I/O,
I tried to use your method after many failed.
It's fine when I tested it on my PC,
but when I pressed Ctrl + C to end the program in Dev-C++
It printed something on the screen before close.
then I found another article
http://online-judge.uva.es/board/viewtopic.php?p=35401
I tried the method that 'emotional blind' san use in his program,
and I got an AC.
I think the hint that 'Sedefcho' san post in that article is also useful, but I don't know if it's really matter because i changed my program lots of times before I found the problem about input.
--
Please forgive my poor english...
190 wa
Posted: Tue Dec 06, 2005 6:31 pm
where am i going wrong or gimma an input that doesn work
input
output
Code: Select all
#include<math.h>
#include<stdio.h>
int main()
{
double con,cenx,ceny,rad,ax,ay,bx,by,cx,cy,mab,mac,p1,q1,r1,p2,q2,r2;
int sx,sy,sc;
while(scanf("%lf %lf %lf %lf %lf %lf",&ax,&ay,&bx,&by,&cx,&cy)==6)
{
sx=0,sy=0,sc=0;
mab=(by-ay)/(bx-ax);
mac=(cy-ay)/(cx-ax);
p1=mab;p2=mac;
q1=1;q2=1;
r1=(ax+bx)/2+mab*(ay+by)/2;
r2=(ax+cx)/2+mac*(ay+cy)/2;
cenx=(r1*p2-r2*p1)/(q1*p2-q2*p1);
ceny=(q2*r1-q1*r2)/(q2*p1-p2*q1);
rad=pow(pow((cenx-ax),2)+pow((ceny-ay),2),0.5);
if(cenx<0)
sx=1;
if(ceny<0)
sy=1;
if((con=pow(cenx,2)+pow(ceny,2)-pow(rad,2))<0)
sc=1;
if(fabs(cenx)<.001)
printf("x^2 ");
else
{
if(sx==0)
printf("(x - %.3lf)^2 + ",cenx);
else
printf("(x + %.3lf)^2 + ",-cenx);
}
if(fabs(ceny)<.001)
printf("y^2 = %.3lf^2",fabs(rad));
else
{
if(sy==0)
printf("(y - %.3lf)^2 = %.3lf^2",ceny,fabs(rad));
else
printf("(y + %.3lf)^2 = %.3lf^2",-ceny,fabs(rad));
}
printf("\nx^2 + y^2 ");
if(sx==0&&fabs(cenx)>.001)
printf("- %.3lfx ",2*fabs(cenx));
if(sx==1&&fabs(cenx)>.001)
printf("+ %.3lfx ",-(2*cenx));
if(sy==0&&fabs(ceny)>.001)
printf("- %.3lfy ",2*fabs(ceny));
if(sy==1&&fabs(ceny)>.001)
printf("+ %.3lfy ",-(2*ceny));
if(sc==1&&fabs(con)>.001)
printf("- %.3lf ",-con);
if(sc==0&&fabs(con)>.001)
printf("+ %.3lf ",fabs(con));
printf("= 0\n\n");
}
}
Code: Select all
7.0 -5.0 -1.0 1.0 0.0 -6.0
1.0 7.0 8.0 6.0 7.0 -2.0
0 0 5 0 2.5 2.5
Code: Select all
(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2
x^2 + y^2 - 6.000x + 4.000y - 12.000 = 0
(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2
x^2 + y^2 - 7.842x - 4.895y - 7.895 = 0
(x - 2.500)^2 + y^2 = 2.500^2
x^2 + y^2 - 5.000x = 0
Posted: Fri Dec 09, 2005 7:56 am
i get the following output for those input from accepted code..
may be it help
GOOD LUCK
Rocky
Code: Select all
(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2
x^2 + y^2 - 6.000x + 4.000y - 12.000 = 0
(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2
x^2 + y^2 - 7.842x - 4.895y - 7.895 = 0
(x - 2.500)^2 + (y + 0.000)^2 = 2.500^2
x^2 + y^2 - 5.000x - 0.000y - 0.000 = 0
GOOD LUCK
Rocky
190-Circle Through Three points -WA-Already read past posts!
Posted: Tue Jan 03, 2006 2:10 pm
Hi everybody!
I'm getting WA in problem 190 - Circle through three points over and over again...
Can someone please check my code, maybe u can find where it's wrong....
I need some critical test cases or something...
For the following input:
my program generates the following output
I already read all the past posts about this problem, sorry for bothering you all
I've also read about the output with zeros, like the third case of my output... some people say it must me made like i did, other people says that it must be without the brackets or zeros.... What's correct?
My code follows
Please help me!
Thank you all and happy new year!
I'm getting WA in problem 190 - Circle through three points over and over again...
Can someone please check my code, maybe u can find where it's wrong....
I need some critical test cases or something...
For the following input:
Code: Select all
7.0 -5.0 -1.0 1.0 0.0 -6.0
1.0 7.0 8.0 6.0 7.0 -2.0
0 0 5 0 2.5 2.5
Code: Select all
(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2
x^2 + y^2 - 6.000x + 4.000y - 12.000 = 0
(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2
x^2 + y^2 - 7.842x - 4.895y - 7.895 = 0
(x - 2.500)^2 + (y + 0.000)^2 = 2.500^2
x^2 + y^2 - 5.000x - 0.000y - 0.000 = 0
I already read all the past posts about this problem, sorry for bothering you all
I've also read about the output with zeros, like the third case of my output... some people say it must me made like i did, other people says that it must be without the brackets or zeros.... What's correct?
My code follows
Code: Select all
#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
class point
{
public:
double x, y;
};
int main()
{
point p1, p2, p3;
double ma, mb, h, k, r, c, d, e;
while(cin>>p1.x)
{
cin>>p1.y>>p2.x>>p2.y>>p3.x>>p3.y;
ma = (p2.y - p1.y)/(p2.x - p1.x);
mb = (p3.y - p2.y)/(p3.x - p2.x);
h = (ma*mb*(p1.y - p3.y) + mb*(p1.x + p2.x) - ma*(p2.x + p3.x))/(2.0*(mb - ma));
k = (-1.0/(ma+0.000001))*(h - ((p1.x + p2.x))/2) + (p1.y + p2.y)/2.0;
r = sqrt((h - p1.x)*(h - p1.x)+(k - p1.y)*(k - p1.y));
//cout<<"c = "<<c<<endl;
//cout<<"d = "<<d<<endl;
if(h >= 0.000001)
printf("(x - %.3f)^2 + (y ", fabs(h));
else
printf("(x + %.3f)^2 + (y ", fabs((-1)*h));
if(k >= 0.000001)
printf("- %.3f)^2 = %.3f^2", fabs(k), fabs(r));
else
printf("+ %.3f)^2 = %.3f^2", fabs((-1)*k),fabs(r));
c = 2.0*h;
d = 2.0*k;
e = (h*h) + (k*k) - (r*r);
if(c >= 0.001){
if(c >= 0.00000001)
printf("\nx^2 + y^2 - %.3fx ", fabs(c));
else
printf("\nx^2 + y^2 + %.3fx ", fabs(c));}
else printf("\nx^2 + y^2 + %.3fx ", fabs(c));
if(d >= 0.001){
if(d >= 0.00000001)
printf("- %.3fy ", fabs(d));
else
printf("+ %.3fy ", fabs(d));}
else if(d == 0.000) printf("- %.3fy ", fabs(d));
else printf("+ %.3fy ", fabs(d));
if(e > 0.00000001)
printf("+ %.3f = 0\n", fabs(e));
else
printf("- %.3f = 0\n", fabs(e));
printf("\n");
}
return 0;
}
Thank you all and happy new year!