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

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(D) The center of the circle is the intersection point of its diameters. Solving the system of equations:$2x + 3y = -1$ $(i)$$3x - y = 4$ (ii)Multiply

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

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(D) The center of the circle is the intersection point of its diameters. Solving the system of equations:$2x + 3y = -1$ $(i)$$3x - y = 4$ (ii)Multiplying (ii) by $3$: $9x - 3y = 12$ (iii)Adding $(i)$ and (iii): $11x = 11 \Rightarrow x = 1$.Substituting $x = 1$ into (ii): $3(1) - y = 4 \Rightarrow y = -1$.Thus,the center $(h, k)$ is $(1, -1)$.Given the circumference $2\pi r = 10\pi$,we find the radius $r = 5$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.$(x - 1)^2 + (y + 1)^2 = 5^2$$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 $3x - y - 4 = 0$ lie along diameters of a circle of circumference $10\pi$,then the equation of the circle is

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