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(B) Let the centre of the circle be $O = (x, y)$. Since the circle passes through $A(-1, 1)$,$B(2, -1)$,and $C(1, 0)$,we have $OA^2 = OB^2 = OC^2 = r^

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$\frac{\sqrt{130}}{2}$

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(B) Let the centre of the circle be $O = (x, y)$. Since the circle passes through $A(-1, 1)$,$B(2, -1)$,and $C(1, 0)$,we have $OA^2 = OB^2 = OC^2 = r^2$.From $OA^2 = OB^2$:$(x + 1)^2 + (y - 1)^2 = (x - 2)^2 + (y + 1)^2$$x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 - 4x + 4 + y^2 + 2y + 1$$6x - 4y = 3$ ... $(i)$From $OA^2 = OC^2$:$(x + 1)^2 + (y - 1)^2 = (x - 1)^2 + y^2$$x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 - 2x + 1 + y^2$$4x - 2y = -1$ ... $(ii)$Multiplying $(ii)$ by $2$,we get $8x - 4y = -2$ ... $(iii)$Subtracting $(i)$ from $(iii)$:$2x = -5 \implies x = -\frac{5}{2}$Substituting $x$ in $(ii)$:$4(-\frac{5}{2}) - 2y = -1 \implies -10 - 2y = -1 \implies 2y = -9 \implies y = -\frac{9}{2}$Centre $O = (-\frac{5}{2}, -\frac{9}{2})$.Radius $r = \sqrt{(x - 1)^2 + (y - 0)^2} = \sqrt{(-\frac{5}{2} - 1)^2 + (-\frac{9}{2})^2} = \sqrt{(-\frac{7}{2})^2 + (-\frac{9}{2})^2} = \sqrt{\frac{49}{4} + \frac{81}{4}} = \sqrt{\frac{130}{4}} = \frac{\sqrt{130}}{2}$.

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 radius of the circle passing through the points $(-1, 1)$,$(2, -1)$,and $(1, 0)$ is

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