Given the circles x^2 + y^2 - 4x - 5 = 0 and | vedclass

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(C) The length of the tangent from a point $P(\alpha, \beta)$ to a circle $S = 0$ is given by $\sqrt{S(\alpha, \beta)}$.For the first circle $S_1: x^2

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$10\alpha - 2\beta + 11 = 0$

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(C) The length of the tangent from a point $P(\alpha, \beta)$ to a circle $S = 0$ is given by $\sqrt{S(\alpha, \beta)}$.For the first circle $S_1: x^2 + y^2 - 4x - 5 = 0$,the length of the tangent is $\sqrt{\alpha^2 + \beta^2 - 4\alpha - 5}$.For the second circle $S_2: x^2 + y^2 + 6x - 2y + 6 = 0$,the length of the tangent is $\sqrt{\alpha^2 + \beta^2 + 6\alpha - 2\beta + 6}$.Since the lengths of the tangents are equal,we have:$\alpha^2 + \beta^2 - 4\alpha - 5 = \alpha^2 + \beta^2 + 6\alpha - 2\beta + 6$Subtracting $\alpha^2 + \beta^2$ from both sides:$-4\alpha - 5 = 6\alpha - 2\beta + 6$Rearranging the terms to one side:$6\alpha + 4\alpha - 2\beta + 6 + 5 = 0$$10\alpha - 2\beta + 11 = 0$.

Given the circles $x^2 + y^2 - 4x - 5 = 0$ and $x^2 + y^2 + 6x - 2y + 6 = 0$. Let $P$ be a point $(\alpha, \beta)$ such that the lengths of the tangents from $P$ to both circles are equal. Then:

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