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(B) Let $A$ and $B$ be the centers of the two circles. Since the line $12x + 5y - 7 = 0$ is tangent to both circles at $(1, -1)$,the line connecting t

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$(13, 4), (-11, -6)$

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(B) Let $A$ and $B$ be the centers of the two circles. Since the line $12x + 5y - 7 = 0$ is tangent to both circles at $(1, -1)$,the line connecting the centers $AB$ is perpendicular to the tangent at $(1, -1)$. The slope of the tangent is $m_t = -\frac{12}{5}$. Therefore,the slope of the line $AB$ is $m_{AB} = -\frac{1}{m_t} = \frac{5}{12}$. Let $	an 	heta = \frac{5}{12}$,so $\cos 	heta = \frac{12}{13}$ and $\sin 	heta = \frac{5}{13}$. The centers $A$ and $B$ lie on the line $AB$ at a distance of $13$ units from the point $(1, -1)$. Using the parametric form,the coordinates are $(1 \pm 13 \cos 	heta, -1 \pm 13 \sin 	heta)$. Substituting the values: $(1 \pm 13 \cdot \frac{12}{13}, -1 \pm 13 \cdot \frac{5}{13}) = (1 \pm 12, -1 \pm 5)$. This gives the two centers as $(1 + 12, -1 + 5) = (13, 4)$ and $(1 - 12, -1 - 5) = (-11, -6)$.

If the equation of the common tangent at the point $(1, -1)$ to the two circles,each of radius $13$,is $12x + 5y - 7 = 0$,then the centers of the two circles are

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