If the lines 3x - 4y - 7 = 0 and 2x - 3y - 5 | vedclass

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(C) The center of the circle is the intersection point of the two diameters.Solving $3x - 4y = 7$ and $2x - 3y = 5$:Multiply the first by $2$ and the

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$x^2 + y^2 - 2x + 2y - 47 = 0$

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(C) The center of the circle is the intersection point of the two diameters.Solving $3x - 4y = 7$ and $2x - 3y = 5$:Multiply the first by $2$ and the second by $3$: $6x - 8y = 14$ and $6x - 9y = 15$.Subtracting the equations: $(6x - 8y) - (6x - 9y) = 14 - 15 \implies y = -1$.Substituting $y = -1$ into $3x - 4y = 7$: $3x - 4(-1) = 7 \implies 3x + 4 = 7 \implies 3x = 3 \implies x = 1$.So,the center $(h, k)$ is $(1, -1)$.Given the area of the circle is $49\pi$,we have $\pi r^2 = 49\pi \implies r^2 = 49 \implies r = 7$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.$(x - 1)^2 + (y + 1)^2 = 7^2$.$x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.$x^2 + y^2 - 2x + 2y + 2 = 49$.$x^2 + y^2 - 2x + 2y - 47 = 0$.

If the lines $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$ are two diameters of a circle with an area of $49\pi$ square units,then the equation of the circle is:

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