Question

find the equation of circle which passes through (1,3) (3,1) and (0,0)​

Answer 1

the equation of the circle is

X²+ Y²+2gx+2fy+c=0 equation 1 we know the value of C=0

firstly, we put the (1, 3) in equation 1

1²+3²+2g×1+2f×3+c=0

10+2g+6f =0 Equation 2

put (3,1) in equation 1

10+6g+2f+c=0 Equation 3

put (0,0) in equation 1

C=0 Equation 4

from equation 2

10+2g+6f =0

2g+6f= -10

2g= -10-6f/

g= -10-6f/ 2

put the value of g in equation 3

10+6g+2f+c=0

6g+2f = -10

6× (-10-6f/2 ) + 2f = -10

-30 -18f+2f = -10

-16f = -10+30

-16f = 20

f = 20+16

f= 36

put the value of F in equation 2

10+2g+6f =0

2g+6f= -10

2g + 6×(36)= -10

2g = -10 - 216

2g = - 226

g = -118

so we put the value of g and f in the equation 1

x² +y² + 2× -118x + 2× 36y =0

so we get the equation is

x² +y² - 236x + 72y =0