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(B) The equation of the line with slope $m = -1$ and $y$-intercept $c = 1$ is $y = -x + 1$,which simplifies to $x + y - 1 = 0$.Let the circle have rad

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$4$

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(B) The equation of the line with slope $m = -1$ and $y$-intercept $c = 1$ is $y = -x + 1$,which simplifies to $x + y - 1 = 0$.Let the circle have radius $r$. Since the circle touches both coordinate axes,its center must be $(\pm r, \pm r)$.The distance from the center $(\pm r, \pm r)$ to the line $x + y - 1 = 0$ must be equal to the radius $r$.Using the distance formula: $\frac{|\pm r \pm r - 1|}{\sqrt{1^2 + 1^2}} = r$.Case $1$: Center $(r, r)$. Then $\frac{|2r - 1|}{\sqrt{2}} = r \implies |2r - 1| = r\sqrt{2}$.$2r - 1 = r\sqrt{2} \implies r(2 - \sqrt{2}) = 1 \implies r = \frac{1}{2 - \sqrt{2}} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$.$2r - 1 = -r\sqrt{2} \implies r(2 + \sqrt{2}) = 1 \implies r = \frac{1}{2 + \sqrt{2}} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{1}{\sqrt{2}}$.Both values of $r$ are positive,giving two circles in the first quadrant.Case $2$: Center $(-r, r)$. Then $\frac{|-r + r - 1|}{\sqrt{2}} = r \implies \frac{|-1|}{\sqrt{2}} = r \implies r = \frac{1}{\sqrt{2}}$.This gives one circle in the second quadrant.Case $3$: Center $(r, -r)$. Then $\frac{|r - r - 1|}{\sqrt{2}} = r \implies \frac{|-1|}{\sqrt{2}} = r \implies r = \frac{1}{\sqrt{2}}$.This gives one circle in the fourth quadrant.Case $4$: Center $(-r, -r)$. Then $\frac{|-2r - 1|}{\sqrt{2}} = r \implies |2r + 1| = r\sqrt{2}$. Since $r > 0$,$2r + 1 > r\sqrt{2}$,so no solution for $r > 0$.Thus,there are $2 + 1 + 1 = 4$ such circles.

Column $I$	Column $II$
$(A)$ Two intersecting circles	$(p)$ have a common tangent
$(B)$ Two mutually external circles	$(q)$ have a common normal
$(C)$ Two circles,one strictly inside the other	$(r)$ do not have a common tangent
$(D)$ Two branches of a hyperbola	$(s)$ do not have a common normal

The number of circles that touch the coordinate axes and the line with slope $-1$ and $y$-intercept $1$ is:

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