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

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(C) Since the lines $2x - 3y = 5$ and $3x - 4y = 7$ are diameters of the circle,their point of intersection is the center $(h, k)$ of the circle.Solvi

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

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(C) Since the lines $2x - 3y = 5$ and $3x - 4y = 7$ are diameters of the circle,their point of intersection is the center $(h, k)$ of the circle.Solving the system of equations:$2x - 3y = 5$  $(i)$$3x - 4y = 7$  $(ii)$Multiplying $(i)$ by $3$ and $(ii)$ by $2$:$6x - 9y = 15$$6x - 8y = 14$Subtracting the equations: $-y = 1 \Rightarrow y = -1$.Substituting $y = -1$ into $(i)$: $2x - 3(-1) = 5$ $\Rightarrow 2x + 3 = 5$ $\Rightarrow 2x = 2$ $\Rightarrow x = 1$.Thus,the center is $(1, -1)$ and the radius $r = 7$.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 $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circle of radius $7$,then the equation of the circle is

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