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(D) Since the circle touches the $X$-axis at $(1, 0)$,the $x$-coordinate of the center is $1$. Since the circle touches the $Y$-axis at $(0, 1)$,the $

Vedclass · Accepted Answer

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

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(D) Since the circle touches the $X$-axis at $(1, 0)$,the $x$-coordinate of the center is $1$. Since the circle touches the $Y$-axis at $(0, 1)$,the $y$-coordinate of the center is $1$. Thus,the center of the circle is $(1, 1)$ and the radius is $r = 1$. The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the values,we get $(x-1)^2 + (y-1)^2 = 1^2$. Expanding this,we have $(x^2 - 2x + 1) + (y^2 - 2y + 1) = 1$. Simplifying,we get $x^2 + y^2 - 2x - 2y + 1 = 0$.

The equation of the circle which touches the $X$-axis and $Y$-axis at the points $(1, 0)$ and $(0, 1)$ respectively is

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