The point (1, 2) lies inside the circle x2 + y2 – 2x + 6y + 1 = 0.
False
Given circle is x2+y2-2x+6y+1=0 or (x-1)2+(y+3)2=32
Centre is C(1,-3) and radius is 3.
Distance of point P (1,2) from centre is 5.
Thus CP>radius
So, point P lies outside the circle.