The equation of a circle touching the coordin | vedclass

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(D) Since the circle touches both coordinate axes,its center is $(g, g)$ and its radius is $|g|$. The equation of the circle is $(x-g)^2 + (y-g)^2 = g

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All of these

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(D) Since the circle touches both coordinate axes,its center is $(g, g)$ and its radius is $|g|$. The equation of the circle is $(x-g)^2 + (y-g)^2 = g^2$,which simplifies to $x^2 + y^2 - 2gx - 2gy + g^2 = 0$.This circle touches the line $x \cos \alpha + y \sin \alpha - 2 = 0$. The perpendicular distance from the center $(g, g)$ to the line must be equal to the radius $|g|$.Thus,$\frac{|g \cos \alpha + g \sin \alpha - 2|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = |g|$.Since $\sqrt{\cos^2 \alpha + \sin^2 \alpha} = 1$,we have $|g(\cos \alpha + \sin \alpha) - 2| = |g|$.This gives two cases: $g(\cos \alpha + \sin \alpha) - 2 = g$ or $g(\cos \alpha + \sin \alpha) - 2 = -g$.Case $1$: $g(\cos \alpha + \sin \alpha - 1) = 2 \implies g = \frac{2}{\cos \alpha + \sin \alpha - 1}$.Case $2$: $g(\cos \alpha + \sin \alpha + 1) = 2 \implies g = \frac{2}{\cos \alpha + \sin \alpha + 1}$.By considering different quadrants for the center $(g, g)$,we can obtain various signs for the coefficients,leading to all the given forms. Thus,all options are correct.

The equation of a circle touching the coordinate axes and the line $x \cos \alpha + y \sin \alpha = 2$ can be:

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