Question

The point (i,i+1) will lie inside circle x^2+y^2-2x+4y=0

Answer 1

Answer:

x2+y2−2x+4y+1=0

x2−2x+y2+4y+1=0

x2−2x+12−12+y2+4y+22−22+1=0

(x−2)2−1+(y+2)2−4+1=0

(x−1)2+(y+2)2=4

Comparing above equation with the standard equation of circle (x−h)2+(y−k)2=r2, we get 

h=1,k=−2

∴ Centre of circle (1,−2)

I HOPE IT'S HELPFUL FOR YOU...