The two circles x^2 + y^2 - 2x - 3 = 0 and x^ | vedclass

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(B) For the first circle $x^2 + y^2 - 2x - 3 = 0$:Center $C_1 = (1, 0)$,Radius $r_1 = \sqrt{1^2 + 0^2 - (-3)} = \sqrt{4} = 2$.For the second circle $x

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They intersect each other

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(B) For the first circle $x^2 + y^2 - 2x - 3 = 0$:Center $C_1 = (1, 0)$,Radius $r_1 = \sqrt{1^2 + 0^2 - (-3)} = \sqrt{4} = 2$.For the second circle $x^2 + y^2 - 4x - 6y - 8 = 0$:Center $C_2 = (2, 3)$,Radius $r_2 = \sqrt{2^2 + 3^2 - (-8)} = \sqrt{4 + 9 + 8} = \sqrt{21}$.The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(2-1)^2 + (3-0)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$.We check the condition for intersection: $|r_1 - r_2| $|2 - \sqrt{21}| \approx |2 - 4.58| = 2.58$.$d = \sqrt{10} \approx 3.16$.$r_1 + r_2 = 2 + \sqrt{21} \approx 6.58$.Since $2.58 Therefore,the circles intersect each other.

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

The two circles $x^2 + y^2 - 2x - 3 = 0$ and $x^2 + y^2 - 4x - 6y - 8 = 0$ are such that:

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