The equation of the circle with center at (2, | vedclass

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(A) Given the circumference of the circle is $10 \pi$ units.We know that the circumference $C = 2 \pi r$,so $2 \pi r = 10 \pi$,which gives $r = 5$.The

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$x^2+y^2-4x+6y-12=0$

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(A) Given the circumference of the circle is $10 \pi$ units.We know that the circumference $C = 2 \pi r$,so $2 \pi r = 10 \pi$,which gives $r = 5$.The center of the circle is $(h, k) = (2, -3)$.The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 2)^2 + (y - (-3))^2 = 5^2$.$(x - 2)^2 + (y + 3)^2 = 25$.Expanding this,we get $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 25$.$x^2 + y^2 - 4x + 6y + 13 = 25$.$x^2 + y^2 - 4x + 6y - 12 = 0$.

The equation of the circle with center at $(2, -3)$ and circumference $10 \pi$ units is

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