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

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(C) The given equation of the circle is $x^2+y^2-4x-6y+11=0$.Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $2g=-4 \Rightarrow g=-2

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$(1,2)$

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(C) The given equation of the circle is $x^2+y^2-4x-6y+11=0$.Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $2g=-4 \Rightarrow g=-2$ and $2f=-6 \Rightarrow f=-3$.The center of the circle is $(-g, -f) = (2, 3)$.Let the other end of the diameter be $(h, k)$.Since the center of the circle is the midpoint of the diameter,we have:$\frac{h+3}{2} = 2$ $\Rightarrow h+3 = 4$ $\Rightarrow h = 1$$\frac{k+4}{2} = 3$ $\Rightarrow k+4 = 6$ $\Rightarrow k = 2$Thus,the other end of the diameter is $(1, 2)$.

If one end of the diameter of the circle $x^2+y^2-4x-6y+11=0$ is $(3,4)$,then the other end of the diameter is

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