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

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

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

Similar Questions

Match the points on the curve $2y^2 = x + 1$ with the slopes of the normals at those points and choose the correct answer.
$A. (7, 2)$ $1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$ $2. -8$
$C. (1, -1)$ $3. 4$
$D. (3, \sqrt{2})$ $4. 0$
$5. -2\sqrt{2}$

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