If the line 4x - 3y + 7 = 0 touches the circl | vedclass

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(C) The equation of the circle is $x^2 + y^2 - 6x + 4y - 12 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3, f = 2, c = -12$. The

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(C) The equation of the circle is $x^2 + y^2 - 6x + 4y - 12 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3, f = 2, c = -12$. The center is $(-g, -f) = (3, -2)$.Since the line $4x - 3y + 7 = 0$ is tangent to the circle at $(\alpha, \beta)$,the radius passing through the point of contact is perpendicular to the tangent.The slope of the tangent is $m_1 = \frac{4}{3}$.The slope of the normal (radius) passing through $(3, -2)$ and $(\alpha, \beta)$ is $m_2 = \frac{\beta - (-2)}{\alpha - 3} = \frac{\beta + 2}{\alpha - 3}$.Since $m_1 	imes m_2 = -1$,we have $\frac{4}{3} 	imes \frac{\beta + 2}{\alpha - 3} = -1$,which implies $4(\beta + 2) = -3(\alpha - 3)$ $\Rightarrow 4\beta + 8 = -3\alpha + 9$ $\Rightarrow 3\alpha + 4\beta = 1$.Also,the point $(\alpha, \beta)$ lies on the line $4\alpha - 3\beta + 7 = 0$.Solving the system:$3\alpha + 4\beta = 1$ $(i)$$4\alpha - 3\beta = -7$ (ii)Multiplying $(i)$ by $3$ and (ii) by $4$: $9\alpha + 12\beta = 3$ and $16\alpha - 12\beta = -28$.Adding these,$25\alpha = -25 \Rightarrow \alpha = -1$.Substituting $\alpha = -1$ into $(i)$: $3(-1) + 4\beta = 1$ $\Rightarrow 4\beta = 4$ $\Rightarrow \beta = 1$.Thus,$\alpha + 2\beta = -1 + 2(1) = 1$.

$List-I$	$List-II$
$(P) \alpha \text{ equals}$	$(1) (-2,4)$
$(Q) r \text{ equals}$	$(2) \sqrt{5}$
$(R) A_1 \text{ equals}$	$(3) (-2,6)$
$(S) B_1 \text{ equals}$	$(4) 5$
	$(5) (2,4)$

If the line $4x - 3y + 7 = 0$ touches the circle $x^2 + y^2 - 6x + 4y - 12 = 0$ at $(\alpha, \beta)$,then $\alpha + 2\beta =$

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