If the circles x^{2}+y^{2}=9 and x^{2}+y^{2}+ | vedclass

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(A) The centers and radii of the given circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ are $C_{1}(0,0), r_{1}=3$ and $C_{2}(-\alpha, -1),

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$\pm \frac{4}{3}$

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(A) The centers and radii of the given circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ are $C_{1}(0,0), r_{1}=3$ and $C_{2}(-\alpha, -1), r_{2}=\sqrt{(-\alpha)^{2}+(-1)^{2}-1} = \sqrt{\alpha^{2}} = |\alpha|$.Since the two circles touch each other internally,the distance between their centers must be equal to the difference of their radii:$C_{1}C_{2} = |r_{1} - r_{2}|$$\sqrt{(-\alpha - 0)^{2} + (-1 - 0)^{2}} = |3 - |\alpha||$$\sqrt{\alpha^{2} + 1} = |3 - |\alpha||$Squaring both sides:$\alpha^{2} + 1 = (3 - |\alpha|)^{2}$$\alpha^{2} + 1 = 9 - 6|\alpha| + \alpha^{2}$$6|\alpha| = 8$$|\alpha| = \frac{4}{3}$$\alpha = \pm \frac{4}{3}$

If the circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ touch each other internally,then $\alpha$ is equal to

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