The equation of the circle which touches the X-axis and Y-axis at the points (1, 0) and (0, 1) respectively is
- $x^2+y^2-4 y+3=0$
- $x^2+y^2-2 y+2=0$
- $x^2+y^2-2 x-2 y+2=0$
- $x^2+y^2-2 x-2 y+1=0$
Solution

$\therefore$ Equation of circle, $(x-1)^2+(y-1)^2=1^2$ $ x^2+y^2-2 x-2 y+1=0 $
Asked in: AP EAMCET 2021 (25 Aug Shift 2)