The equation of the circle in the first quadr | vedclass

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(A) Since the circle is in the first quadrant and touches both coordinate axes at a distance of $1$ unit from the origin,the center of the circle is $

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Since the circle is in the first quadrant and touches both coordinate axes at a distance of $1$ unit from the origin,the center of the circle is $(1, 1)$ and the radius $r = 1$.The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $h = 1, k = 1$,and $r = 1$:$(x - 1)^2 + (y - 1)^2 = 1^2$$x^2 - 2x + 1 + y^2 - 2y + 1 = 1$$x^2 + y^2 - 2x - 2y + 1 = 0$.

The equation of the circle in the first quadrant touching each coordinate axis at a distance of one unit from the origin is

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