If the circles x^2 + y^2 = 16 and x^2 + y^2 + | vedclass

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(D) The first circle is $x^2 + y^2 = 16$, which has center $C_1 = (0, 0)$ and radius $r_1 = 4$.The second circle is $x^2 + y^2 + 2ax + 4y + 4 = 0$. Re

Vedclass · Accepted Answer

$-3$

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(D) The first circle is $x^2 + y^2 = 16$, which has center $C_1 = (0, 0)$ and radius $r_1 = 4$.The second circle is $x^2 + y^2 + 2ax + 4y + 4 = 0$. Rewriting it as $(x+a)^2 + (y+2)^2 = a^2 + 4 - 4 = a^2$, we get center $C_2 = (-a, -2)$ and radius $r_2 = |a|$.For two circles to touch internally, the distance between their centers $d = C_1C_2$ must be equal to the difference of their radii: $d = |r_1 - r_2|$.$d = \sqrt{(-a - 0)^2 + (-2 - 0)^2} = \sqrt{a^2 + 4}$.So, $\sqrt{a^2 + 4} = |4 - |a||$.Squaring both sides: $a^2 + 4 = (4 - |a|)^2 = 16 - 8|a| + a^2$.$4 = 16 - 8|a| \implies 8|a| = 12 \implies |a| = 12/8 = 3/2$.Thus, $a = \pm 3/2$. Given the options, we select the value that satisfies the condition.

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

If the circles $x^2 + y^2 = 16$ and $x^2 + y^2 + 2ax + 4y + 4 = 0$ touch each other internally, then $a =$

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