If one end of a 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

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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 $2g = -4 \implies g = -2$ and $2f = -6 \implies f = -3$.The centre of the circle is $(-g, -f) = (2, 3)$.Let the given end of the diameter be $A = (3, 4)$ and the other end be $B = (x, y)$.Since the centre $C(2, 3)$ is the midpoint of the diameter $AB$,we have:$\frac{3 + x}{2} = 2 \implies 3 + x = 4 \implies x = 1$$\frac{4 + y}{2} = 3 \implies 4 + y = 6 \implies y = 2$Thus,the other end is $(1, 2)$.

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

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