In List-I,a pair of circles is given in A,B,C | vedclass

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(A) The angle $	heta$ between two circles with radii $r_1, r_2$ and distance between centers $d$ is given by $\cos 	heta = \frac{r_1^2+r_2^2-d^2}{2

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$A-II, B-I, C-III$

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(A) The angle $	heta$ between two circles with radii $r_1, r_2$ and distance between centers $d$ is given by $\cos 	heta = \frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}$.$(A)$ $(x-2)^2+y^2=2$ $(r_1=\sqrt{2}, c_1=(2,0))$ and $(x-2)^2+(y-1)^2=1$ $(r_2=1, c_2=(2,1))$.$d^2 = (2-2)^2 + (1-0)^2 = 1$.$\cos 	heta = \frac{2+1-1}{2 	imes \sqrt{2} 	imes 1} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.$	heta = 45^{\circ}$ or $135^{\circ}$. Thus,$(A)$ matches with $II$.$(B)$ $x^2+y^2-6x-6y+9=0$ $(r_1=\sqrt{3^2+3^2-9}=3, c_1=(3,3))$ and $x^2+y^2-4x+4y-9=0$ $(r_2=\sqrt{2^2+(-2)^2-(-9)}=\sqrt{17}, c_2=(2,-2))$.$d^2 = (3-2)^2 + (3-(-2))^2 = 1^2 + 5^2 = 26$.$\cos 	heta = \frac{9+17-26}{2 	imes 3 	imes \sqrt{17}} = 0$.$	heta = 90^{\circ}$. Thus,$(B)$ matches with $I$.$(C)$ $x^2+y^2+4x-14y+28=0$ $(r_1=\sqrt{(-2)^2+7^2-28}=5, c_1=(-2,7))$ and $x^2+y^2+4x-5=0$ $(r_2=\sqrt{(-2)^2+0^2-(-5)}=3, c_2=(-2,0))$.$d^2 = (-2-(-2))^2 + (7-0)^2 = 49$.$\cos 	heta = \frac{25+9-49}{2 	imes 5 	imes 3} = \frac{-15}{30} = -\frac{1}{2}$.$	heta = 120^{\circ}$ or $60^{\circ}$. Thus,$(C)$ matches with $III$.Therefore,the correct matching is $A-II, B-I, C-III$.

List-$I$	List-$II$
$(A)$ $(x-2)^2+y^2=2$,$(x-2)^2+(y-1)^2=1$	$I.$ $90^{\circ}$
$(B)$ $x^2+y^2-6x-6y+9=0$,$x^2+y^2-4x+4y-9=0$	$II.$ $135^{\circ}$
$(C)$ $x^2+y^2+4x-14y+28=0$,$x^2+y^2+4x-5=0$	$III.$ $60^{\circ}$
	$IV.$ $30^{\circ}$

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