If the circles x^2 + y^2 - 9 = 0 and x^2 + y^ | vedclass

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(D) The first circle is $x^2 + y^2 = 3^2$,with center $C_1 = (0, 0)$ and radius $r_1 = 3$.The second circle is $x^2 + y^2 + 2ax + 2y + 1 = 0$,with cen

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Both $A$ and $B$

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(D) The first circle is $x^2 + y^2 = 3^2$,with center $C_1 = (0, 0)$ and radius $r_1 = 3$.The second circle is $x^2 + y^2 + 2ax + 2y + 1 = 0$,with center $C_2 = (-a, -1)$ and radius $r_2 = \sqrt{(-a)^2 + (-1)^2 - 1} = \sqrt{a^2}$. Since $r_2$ must be positive,$r_2 = |a|$.The distance between centers is $d = \sqrt{(-a - 0)^2 + (-1 - 0)^2} = \sqrt{a^2 + 1}$.Case $1$: Circles touch externally,$d = r_1 + r_2$.$\sqrt{a^2 + 1} = 3 + |a|$. Squaring both sides: $a^2 + 1 = 9 + a^2 + 6|a| \Rightarrow 6|a| = -8$,which has no real solution.Case $2$: Circles touch internally,$d = |r_1 - r_2|$.$\sqrt{a^2 + 1} = |3 - |a||$. Squaring both sides: $a^2 + 1 = 9 + a^2 - 6|a|$ $\Rightarrow 6|a| = 8$ $\Rightarrow |a| = 4/3$.Thus,$a = 4/3$ or $a = -4/3$.

If the circles $x^2 + y^2 - 9 = 0$ and $x^2 + y^2 + 2ax + 2y + 1 = 0$ touch each other,then $a =$

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