Answer:
Step-by-step explanation:
let the eq. of circle be a(x²+y²)+bx+cy+d=0
x²+y²+b/a x+c/a y+d/a=0
if it passes through (0,0)
0²+0²+b/a*0+c/a*0+d/a=0
so d=0
Hence x²+y²+b/a x+c/a y=0
if it passes through (1,2) and (3,-1)
1²+2²+b/a×1+c/a×2=0
5+b/a+2c/a=0 ...(1)
and 3²+(-1)²+b/a*3+c/a*(-1)=0
10+3b/a-c/a=0 ...(2)
(1)+2(2) gives
5+b/a+2 c/a+20+6 b/a -2 c/a=0
7 b/a+25=0
b/a=-25/7
from (1)
5-25/7+2 c/a=0
2 c/a+10/7=0
2 c/a=-10/7
c/a=-5/7
so eq. of circle is x²+y²-25/7 x-5/7 y=0
7(x²+y²)-25 x-5 y=0
