A circle touches both axes and its center lie | vedclass

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(A) Since the circle touches both axes and lies in the fourth quadrant with radius $r = 1$,the center of the circle must be $(1, -1)$.The standard equ

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$x^2 + y^2 - 2x + 2y + 1 = 0$

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(A) Since the circle touches both axes and lies in the fourth quadrant with radius $r = 1$,the center of the circle must be $(1, -1)$.The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $h = 1$,$k = -1$,and $r = 1$:$(x - 1)^2 + (y - (-1))^2 = 1^2$$(x - 1)^2 + (y + 1)^2 = 1$Expanding the equation:$(x^2 - 2x + 1) + (y^2 + 2y + 1) = 1$$x^2 + y^2 - 2x + 2y + 2 = 1$$x^2 + y^2 - 2x + 2y + 1 = 0$

$A$ circle touches both axes and its center lies in the fourth quadrant. If its radius is $1$,then its equation is:

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