Answer

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**Hint:**We first try to form the given circle in its general form of $ {{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}} $ to find the centre and the radius. from there we place the values of the y coordinate of the centre as 0. Then we put the value of $ y=0 $ in the equation $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ to get end points of the diameter. We put the possible given options and check the most appropriate one.

**Complete step by step answer:**

The equation $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ represents a circle with X-axis as a diameter and radius a.

As the diameter is a part of X-axis, it means that the centre is on the X-axis and the y coordinate of the centre is 0.

It’s given that the equation of the circle is $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ . We transform it in its general form of $ {{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}} $ and get $ {{\left( x-g \right)}^{2}}+{{\left( y-f \right)}^{2}}={{\left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)}^{2}} $ . O is the centre.

Equating with the general equation of circle $ {{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}} $ , we get the centre as $ O\equiv \left( -g,-f \right) $ and the radius as $ \sqrt{{{g}^{2}}+{{f}^{2}}-c} $ units.

We can put the value $ -f=0 $ which means $ f=0 $ . Also, the radius is a.

This means the centre is $ O\equiv \left( -g,0 \right) $ and the radius as $ \sqrt{{{g}^{2}}+{{f}^{2}}-c}=a $ units.

Solving the radius, we get $ {{g}^{2}}-c={{a}^{2}} $ .

Now we put the value of $ y=0 $ in the equation $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ to get the two end points of the diameter. This gives $ {{x}^{2}}+2gx+c=0 $ . Solving we get

$ x=\dfrac{-2g\pm \sqrt{4{{g}^{2}}-4c}}{2}=-g\pm \sqrt{{{g}^{2}}-c} $ .

Now we place the given two options which are possible and they are $ f=0,g=a,c=3{{a}^{2}} $ and $ f=0,g=-2a,c=3{{a}^{2}} $ . In both cases $ f=0,c=3{{a}^{2}} $ . Now we put $ c=3{{a}^{2}} $ in $ {{g}^{2}}-c={{a}^{2}} $ .

We get $ {{g}^{2}}=4{{a}^{2}} $ which gives $ g=\pm 2a $ .

**Therefore, the correct option is C**.

**Note:**

We need to remember that the general equation $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ always makes an intercept $ 2\sqrt{{{g}^{2}}-c} $ units on the X-axis and $ 2\sqrt{{{f}^{2}}-c} $ units on the Y-axis. We use that to find the diameter length in the circle of $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ .

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