If the lines 2x + 3y + 1 = 0 and 3x - y - 4 = | vedclass

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(A) The center of the circle is the intersection of the two diameters $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$.Multiplying the second equation by $3$,we

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

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(A) The center of the circle is the intersection of the two diameters $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$.Multiplying the second equation by $3$,we get $9x - 3y - 12 = 0$.Adding this to the first equation: $(2x + 3y + 1) + (9x - 3y - 12) = 0 \implies 11x - 11 = 0 \implies x = 1$.Substituting $x = 1$ into $3x - y - 4 = 0$,we get $3(1) - y - 4 = 0 \implies y = -1$.So,the center $(h, k) = (1, -1)$.The circumference is $2\pi r = 10\pi$,which gives $r = 5$.The equation of the circle 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 the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$ are diameters of a circle with circumference $10\pi$,find the equation of the circle.

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