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(C) Let the new coordinates be $(x', y')$ such that $x = x' + h$ and $y = y' + k$. Substituting these into the given equation $x^2 + y^2 - 4x + 6y - 7

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$(2, -3)$

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(C) Let the new coordinates be $(x', y')$ such that $x = x' + h$ and $y = y' + k$. Substituting these into the given equation $x^2 + y^2 - 4x + 6y - 7 = 0$ gives:$(x' + h)^2 + (y' + k)^2 - 4(x' + h) + 6(y' + k) - 7 = 0$Expanding the terms:$x'^2 + 2hx' + h^2 + y'^2 + 2ky' + k^2 - 4x' - 4h + 6y' + 6k - 7 = 0$Grouping the terms:$x'^2 + y'^2 + x'(2h - 4) + y'(2k + 6) + (h^2 + k^2 - 4h + 6k - 7) = 0$To eliminate the linear terms,the coefficients of $x'$ and $y'$ must be zero:$2h - 4 = 0 \Rightarrow h = 2$$2k + 6 = 0 \Rightarrow k = -3$Thus,the point $(h, k)$ is $(2, -3)$.

Without changing the direction of the coordinate axes,the origin is shifted to $(h, k)$ such that the linear (first-degree) terms in the equation $x^2 + y^2 - 4x + 6y - 7 = 0$ are eliminated. Then the point $(h, k)$ is:

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