If the circles x^2+y^2-8x-8y+28=0 and x^2+y^2 | vedclass

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(C) For the first circle $S_1: x^2+y^2-8x-8y+28=0$,the center $C_1 = (4, 4)$ and radius $r_1 = \sqrt{4^2+4^2-28} = \sqrt{16+16-28} = \sqrt{4} = 2$. Fo

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$\alpha=1$

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(C) For the first circle $S_1: x^2+y^2-8x-8y+28=0$,the center $C_1 = (4, 4)$ and radius $r_1 = \sqrt{4^2+4^2-28} = \sqrt{16+16-28} = \sqrt{4} = 2$. For the second circle $S_2: x^2+y^2-8x-6y+25-\alpha^2=0$,the center $C_2 = (4, 3)$ and radius $r_2 = \sqrt{4^2+3^2-(25-\alpha^2)} = \sqrt{16+9-25+\alpha^2} = \sqrt{\alpha^2} = |\alpha|$. Two circles have only one common tangent if they touch each other internally or externally. The distance between centers $C_1 C_2 = \sqrt{(4-4)^2 + (4-3)^2} = \sqrt{0^2 + 1^2} = 1$. For internal touch,$|r_1 - r_2| = C_1 C_2 \Rightarrow |2 - |\alpha|| = 1$. This gives $2 - |\alpha| = 1$ $\Rightarrow |\alpha| = 1$ $\Rightarrow \alpha = \pm 1$ or $2 - |\alpha| = -1$ $\Rightarrow |\alpha| = 3$ $\Rightarrow \alpha = \pm 3$. Given the options,$\alpha = 1$ is the correct value.

If the circles $x^2+y^2-8x-8y+28=0$ and $x^2+y^2-8x-6y+25-\alpha^2=0$ have only one common tangent,then $\alpha=$

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