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(C) The given equation of the circle is $x^2+y^2+4x-8y+5=0$.Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$, we get $2g=4 \Rightarrow g

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$(-6,7)$

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(C) The given equation of the circle is $x^2+y^2+4x-8y+5=0$.Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$, we get $2g=4 \Rightarrow g=2$ and $2f=-8 \Rightarrow f=-4$.The center of the circle is $(-g, -f) = (-2, 4)$.Let the coordinates of the other end of the diameter be $(h, k)$.Since the center of the circle is the midpoint of the diameter, we have:$\frac{h+2}{2} = -2$ $\Rightarrow h+2 = -4$ $\Rightarrow h = -6$$\frac{k+1}{2} = 4$ $\Rightarrow k+1 = 8$ $\Rightarrow k = 7$Thus, the coordinates of the other end are $(-6, 7)$.

If the coordinates of one end of a diameter of the circle $x^2+y^2+4x-8y+5=0$ are $(2,1)$, then the coordinates of the other end are:

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