Answer:
Step-by-step explanation:
let the eq. of circle be x²+y²+2gx+2fy+c=0 ...(1)
∵ it passes through (-1,2),(4,2),(-3,4)
∴ (-1)²+2²-2g+4f+c=0
-2g+4f+c=-5 ...(2)
4²+2²+8g+4f+c=0
8g+4f+c=-20 ...(3)
(-3)²+4²-6g+8f+c=0
-6g+8f+c=-25 ...(4)
(3)-(2) gives
10 g=15
2g=3
g=3/2
(3)-(4) gives
14 g-4f=5
14×(3/2)-4f=5
-4f=5-21=-16
f=4
from (2)
-2(3/2)+4×4+c=-5
-3+16+c=-5
c=-5-13
c=-18
so eq. of circle is
x²+y²+2(3/2)x+2(4)y-18=0
x²+y²+3x+8y-18=0
