If one end of a diameter of the circle 3x^{2} | vedclass

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(C) The given equation of the circle is $3x^{2}+3y^{2}-9x+6y+5=0$. Dividing by $3$, we get $x^{2}+y^{2}-3x+2y+\frac{5}{3}=0$. The center of the circle

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$(2,-4)$

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(C) The given equation of the circle is $3x^{2}+3y^{2}-9x+6y+5=0$. Dividing by $3$, we get $x^{2}+y^{2}-3x+2y+\frac{5}{3}=0$. The center of the circle $(h, k)$ is given by $(-\frac{g}{2}, -f)$, where $2g = -3$ and $2f = 2$. Thus, the center is $(\frac{3}{2}, -1)$. We know that the center of the circle is the midpoint of the diameter. Let the given end of the diameter be $(x_1, y_1) = (1, 2)$ and the other end be $(x_2, y_2) = (h, k)$. Using the midpoint formula: $\frac{1+h}{2} = \frac{3}{2}$ $\Rightarrow 1+h = 3$ $\Rightarrow h = 2$. $\frac{2+k}{2} = -1$ $\Rightarrow 2+k = -2$ $\Rightarrow k = -4$. Therefore, the other end of the diameter is $(2, -4)$.

If one end of a diameter of the circle $3x^{2}+3y^{2}-9x+6y+5=0$ is $(1,2)$, then the other end is

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