If the equation of the polar of the point (al | vedclass

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(A) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.

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$-5$

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(A) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. Given the circle $x^2+y^2-4x-6y-12=0$,we have $g=-2, f=-3, c=-12$. The point is $(\alpha, -1)$. Substituting these into the formula: $x(\alpha)+y(-1)-2(x+\alpha)-3(y-1)-12=0$. Simplifying: $\alpha x - y - 2x - 2\alpha - 3y + 3 - 12 = 0$. $(\alpha-2)x - 4y - 2\alpha - 9 = 0$. Given that this equation is $y=\beta$,or $0x + y - \beta = 0$. Comparing the coefficients of $x$ and $y$: For $x$: $\alpha-2 = 0 \implies \alpha = 2$. For $y$: $\frac{-4}{1} = \frac{-2\alpha-9}{-\beta}$. $-4 = \frac{-2(2)-9}{-\beta} \implies -4 = \frac{-13}{-\beta} \implies -4 = \frac{13}{\beta}$. $\beta = -\frac{13}{4}$. Now,$4(\alpha+\beta) = 4(2 - \frac{13}{4}) = 4(\frac{8-13}{4}) = 4(-\frac{5}{4}) = -5$.

If the equation of the polar of the point $(\alpha, -1)$ with respect to the circle $x^2+y^2-4x-6y-12=0$ is $y=\beta$,then $4(\alpha+\beta)=$

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