If one end of a diameter of the circle x^2 + | vedclass

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(A) The given equation of the circle is $x^2 + y^2 - 3x + 8y - 4 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g

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$(-3, -5)$

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(A) The given equation of the circle is $x^2 + y^2 - 3x + 8y - 4 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -3 \Rightarrow g = -\frac{3}{2}$ and $2f = 8 \Rightarrow f = 4$.The center of the circle is $(-g, -f) = (\frac{3}{2}, -4)$.Let the other end of the diameter be $(h, k)$.Since the center is the midpoint of the diameter,we have:$\frac{6 + h}{2} = \frac{3}{2}$ $\Rightarrow 6 + h = 3$ $\Rightarrow h = -3$.$\frac{-3 + k}{2} = -4$ $\Rightarrow -3 + k = -8$ $\Rightarrow k = -5$.Thus,the other end of the diameter is $(-3, -5)$.

If one end of a diameter of the circle $x^2 + y^2 - 3x + 8y - 4 = 0$ is $(6, -3)$,then what is the other end of the diameter?

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