If the line 3x - 4y = 1 touches the circle (x | vedclass

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(B) Since $(\alpha, \beta)$ lies on the line $3x - 4y = 1$,we have $3\alpha - 4\beta = 1$,which implies $\beta = \frac{3\alpha - 1}{4}$.Since $(\alpha

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$\alpha = -\frac{1}{5}, \beta = -\frac{2}{5}$

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(B) Since $(\alpha, \beta)$ lies on the line $3x - 4y = 1$,we have $3\alpha - 4\beta = 1$,which implies $\beta = \frac{3\alpha - 1}{4}$.Since $(\alpha, \beta)$ also lies on the circle $(x - 1)^2 + (y + 2)^2 = 4$,we substitute $\beta$:$(\alpha - 1)^2 + (\frac{3\alpha - 1}{4} + 2)^2 = 4$$(\alpha - 1)^2 + (\frac{3\alpha + 7}{4})^2 = 4$$16(\alpha - 1)^2 + (3\alpha + 7)^2 = 64$$16(\alpha^2 - 2\alpha + 1) + (9\alpha^2 + 42\alpha + 49) = 64$$16\alpha^2 - 32\alpha + 16 + 9\alpha^2 + 42\alpha + 49 = 64$$25\alpha^2 + 10\alpha + 1 = 0$$(5\alpha + 1)^2 = 0$$\alpha = -\frac{1}{5}$.Substituting $\alpha$ into the line equation: $\beta = \frac{3(-\frac{1}{5}) - 1}{4} = \frac{-\frac{3}{5} - 1}{4} = \frac{-\frac{8}{5}}{4} = -\frac{2}{5}$.Thus,$(\alpha, \beta) = (-\frac{1}{5}, -\frac{2}{5})$.

If the line $3x - 4y = 1$ touches the circle $(x - 1)^2 + (y + 2)^2 = 4$ at $(\alpha, \beta)$,the values of $\alpha$ and $\beta$ are

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