The equation of the circle touching x = 0,y = | vedclass

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(D) The circle touches the lines $x = 0$ (y-axis),$y = 0$ (x-axis),and $x = 4$.Since it touches $x = 0$ and $x = 4$,the diameter of the circle is the

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$x^2 + y^2 - 4x - 4y + 4 = 0$

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(D) The circle touches the lines $x = 0$ (y-axis),$y = 0$ (x-axis),and $x = 4$.Since it touches $x = 0$ and $x = 4$,the diameter of the circle is the distance between these lines,which is $4 - 0 = 4$. Thus,the radius $r = 2$.The center of the circle must be at a distance $r = 2$ from both $x = 0$ and $y = 0$,so the center is $(2, 2)$.The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $h = 2, k = 2, r = 2$,we get $(x - 2)^2 + (y - 2)^2 = 2^2$.$(x^2 - 4x + 4) + (y^2 - 4y + 4) = 4$.$x^2 + y^2 - 4x - 4y + 4 = 0$.

The equation of the circle touching $x = 0$,$y = 0$ and $x = 4$ is

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