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(B) The circumference of the circle is $10 \pi$. Given $2 \pi r = 10 \pi$,we find the radius $r = 5$. Since the diameters of a circle intersect at its

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$x^2 + y^2 - 2x + 2y - 23 = 0$

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(B) The circumference of the circle is $10 \pi$. Given $2 \pi r = 10 \pi$,we find the radius $r = 5$. Since the diameters of a circle intersect at its center,we solve the system of equations: $2x + 3y + 1 = 0$ $(i)$ $3x - y - 4 = 0$ $(ii)$ From $(ii)$,$y = 3x - 4$. Substituting this into $(i)$: $2x + 3(3x - 4) + 1 = 0$ $2x + 9x - 12 + 1 = 0$ $11x - 11 = 0 \Rightarrow x = 1$. Then $y = 3(1) - 4 = -1$. The center of the circle is $(1, -1)$. The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$. $(x - 1)^2 + (y + 1)^2 = 5^2$ $x^2 - 2x + 1 + y^2 + 2y + 1 = 25$ $x^2 + y^2 - 2x + 2y - 23 = 0$.

If two diameters of a circle of circumference $10 \pi$ lie along the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$,then the equation of the circle is

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