In List-I,each item contains equations of two | vedclass

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(A) . For $x^2+y^2+2x+8y-23=0$,center $C_1(-1,-4)$,radius $r_1=\sqrt{1+16+23}=\sqrt{40}=2\sqrt{10}$. For $x^2+y^2-4x-10y+19=0$,center $C_2(2,5)$,radiu

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$A-IV, B-V, C-III, D-II$

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(A) . For $x^2+y^2+2x+8y-23=0$,center $C_1(-1,-4)$,radius $r_1=\sqrt{1+16+23}=\sqrt{40}=2\sqrt{10}$. For $x^2+y^2-4x-10y+19=0$,center $C_2(2,5)$,radius $r_2=\sqrt{4+25-19}=\sqrt{10}$. Distance $C_1C_2=\sqrt{(2-(-1))^2+(5-(-4))^2}=\sqrt{3^2+9^2}=\sqrt{90}=3\sqrt{10}$. Since $C_1C_2=r_1+r_2$,the circles touch externally,so there are $3$ common tangents.$B$. For $x^2+y^2=1$,center $C_1(0,0)$,radius $r_1=1$. For $x^2+y^2-2x-6y+6=0$,center $C_2(1,3)$,radius $r_2=\sqrt{1+9-6}=2$. Distance $C_1C_2=\sqrt{1^2+3^2}=\sqrt{10}$. Since $r_1+r_2=3  r_1+r_2$ $(\sqrt{10} > 3)$,the circles are separate,so there are $4$ common tangents.$C$. For $x^2+y^2-8x+2y=0$,center $C_1(4,-1)$,radius $r_1=\sqrt{16+1}=\sqrt{17}$. For $x^2+y^2-2x-16y+25=0$,center $C_2(1,8)$,radius $r_2=\sqrt{1+64-25}=\sqrt{40}=2\sqrt{10}$. Distance $C_1C_2=\sqrt{(4-1)^2+(-1-8)^2}=\sqrt{3^2+(-9)^2}=\sqrt{90}=3\sqrt{10}$. Since $|r_1-r_2| $D$. For $x^2+y^2=4$,center $C_1(0,0)$,radius $r_1=2$. For $x^2+y^2-2x=0$,center $C_2(1,0)$,radius $r_2=1$. Distance $C_1C_2=\sqrt{1^2+0^2}=1$. Since $C_1C_2=|r_1-r_2|=1$,the circles touch internally,so there is $1$ common tangent.Thus,the correct match is $A-IV, B-V, C-III, D-II$.

List-$I$	List-$II$
$A$. $x^2+y^2+2x+8y-23=0$,$x^2+y^2-4x-10y+19=0$	$I$. $0$
$B$. $x^2+y^2=1$,$x^2+y^2-2x-6y+6=0$	$II$. $1$
$C$. $x^2+y^2-8x+2y=0$,$x^2+y^2-2x-16y+25=0$	$III$. $2$
$D$. $x^2+y^2=4$,$x^2+y^2-2x=0$	$IV$. $3$
	$V$. $4$

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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