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(A) The center of the circle is the intersection of the two diameters $2x - 3y = 5$ and $3x - 4y = 7$.Multiplying the first equation by $4$ and the se

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$2$

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(A) The center of the circle is the intersection of the two diameters $2x - 3y = 5$ and $3x - 4y = 7$.Multiplying the first equation by $4$ and the second by $3$,we get $8x - 12y = 20$ and $9x - 12y = 21$.Subtracting the first from the second gives $x = 1$. Substituting $x = 1$ into $2(1) - 3y = 5$,we get $-3y = 3$,so $y = -1$. Thus,the center $C$ is $(1, -1)$.The line $AB$ passes through $A\left(-\frac{22}{7}, -4\right)$ and $B\left(-\frac{1}{7}, 3\right)$. The slope $m$ of $AB$ is $\frac{3 - (-4)}{-1/7 - (-22/7)} = \frac{7}{21/7} = \frac{7}{3} = \frac{7}{3}$.The equation of line $AB$ is $y - 3 = \frac{7}{3}(x + \frac{1}{7})$,which simplifies to $3y - 9 = 7x + 1$,or $7x - 3y + 10 = 0$.Since the line intersects the circle at only one point $P$,it is a tangent to the circle at $P$. Thus,$CP$ is perpendicular to $AB$.The slope of $CP$ is $-\frac{1}{7/3} = -\frac{3}{7}$. The equation of line $CP$ passing through $C(1, -1)$ is $y + 1 = -\frac{3}{7}(x - 1)$,which simplifies to $7y + 7 = -3x + 3$,or $3x + 7y + 4 = 0$.Solving the system $7x - 3y = -10$ and $3x + 7y = -4$:Multiply the first by $7$ and the second by $3$: $49x - 21y = -70$ and $9x + 21y = -12$.Adding these gives $58x = -82$,so $x = \alpha = -\frac{41}{29}$.Substituting $x = -\frac{41}{29}$ into $3x + 7y = -4$: $3(-\frac{41}{29}) + 7y = -4 \implies -\frac{123}{29} + 7y = -4 \implies 7y = -4 + \frac{123}{29} = \frac{-116 + 123}{29} = \frac{7}{29} \implies y = \beta = \frac{1}{29}$.Finally,$17\beta - \alpha = 17(\frac{1}{29}) - (-\frac{41}{29}) = \frac{17 + 41}{29} = \frac{58}{29} = 2$.

The equations of two diameters of a circle are $2x - 3y = 5$ and $3x - 4y = 7$. The line joining the points $\left(-\frac{22}{7}, -4\right)$ and $\left(-\frac{1}{7}, 3\right)$ intersects the circle at only one point $P(\alpha, \beta)$. Then $17\beta - \alpha$ is equal to

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