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(C) To find the center $(h, k)$,solve the system of equations:$2x - 3y = -4$ $(i)$$3x + 4y = 5$ (ii)Multiplying $(i)$ by $4$ and (ii) by $3$: $8x - 12

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$17(x^{2} + y^{2}) + 2x - 44y = 0$

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(C) To find the center $(h, k)$,solve the system of equations:$2x - 3y = -4$ $(i)$$3x + 4y = 5$ (ii)Multiplying $(i)$ by $4$ and (ii) by $3$: $8x - 12y = -16$ and $9x + 12y = 15$.Adding them gives $17x = -1$,so $x = -\frac{1}{17}$.Substituting $x$ into $(i)$: $2(-\frac{1}{17}) - 3y = -4 \implies -3y = -4 + \frac{2}{17} = -\frac{66}{17} \implies y = \frac{22}{17}$.Center is $(h, k) = (-\frac{1}{17}, \frac{22}{17})$.Since the circle passes through the origin $(0, 0)$,the radius $r$ is the distance from the center to the origin:$r^{2} = h^{2} + k^{2} = (-\frac{1}{17})^{2} + (\frac{22}{17})^{2} = \frac{1 + 484}{289} = \frac{485}{289}$.The equation of the circle is $(x - h)^{2} + (y - k)^{2} = r^{2}$:$(x + \frac{1}{17})^{2} + (y - \frac{22}{17})^{2} = \frac{485}{289}$$x^{2} + \frac{2x}{17} + \frac{1}{289} + y^{2} - \frac{44y}{17} + \frac{484}{289} = \frac{485}{289}$$x^{2} + y^{2} + \frac{2x}{17} - \frac{44y}{17} + \frac{485}{289} = \frac{485}{289}$$x^{2} + y^{2} + \frac{2x}{17} - \frac{44y}{17} = 0$Multiplying by $17$: $17(x^{2} + y^{2}) + 2x - 44y = 0$.

Find the equation of the circle passing through the origin and whose center is the intersection point of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$.

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