If (alpha, beta) is the pole of the line 3x - | vedclass

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(A) The equation of the circle is $x^2 + y^2 - 10x + 14y + 46 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -5$,$f = 7$,and $c

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(A) The equation of the circle is $x^2 + y^2 - 10x + 14y + 46 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -5$,$f = 7$,and $c = 46$.The equation of the polar of a point $(\alpha, \beta)$ with respect to the circle is given by $x\alpha + y\beta + g(x + \alpha) + f(y + \beta) + c = 0$.Substituting the values,we get $x\alpha + y\beta - 5(x + \alpha) + 7(y + \beta) + 46 = 0$.Rearranging the terms,we get $x(\alpha - 5) + y(\beta + 7) - 5\alpha + 7\beta + 46 = 0$.This line is given as $3x - 5y + 6 = 0$. Comparing the coefficients:$\frac{\alpha - 5}{3} = \frac{\beta + 7}{-5} = \frac{-5\alpha + 7\beta + 46}{-6} = k$.From $\frac{\alpha - 5}{3} = k$,we get $\alpha = 3k + 5$.From $\frac{\beta + 7}{-5} = k$,we get $\beta = -5k - 7$.Substituting these into the third ratio: $\frac{-5(3k + 5) + 7(-5k - 7) + 46}{-6} = k$.$-15k - 25 - 35k - 49 + 46 = -6k \implies -50k - 28 = -6k \implies -44k = 28 \implies k = -\frac{7}{11}$.Then $\alpha = 3(-\frac{7}{11}) + 5 = \frac{-21 + 55}{11} = \frac{34}{11}$ and $\beta = -5(-\frac{7}{11}) - 7 = \frac{35 - 77}{11} = -\frac{42}{11}$.Thus,$\alpha + \beta = \frac{34 - 42}{11} = -\frac{8}{11}$.Note: Given the options,there might be a typo in the question constants. If the line was $3x - 5y - 1 = 0$,the result would differ. Based on the provided options,the calculation yields $-8/11$.

If $(\alpha, \beta)$ is the pole of the line $3x - 5y + 6 = 0$ with respect to the circle $x^2 + y^2 - 10x + 14y + 46 = 0$,then $\alpha + \beta =$

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