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# In the equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ represents a circle with X-axis as a diameter and radius a, then which of the following is possible.A. $f=2a,g=0,c=3{{a}^{2}}$ B. $f=0,g=a,c=3{{a}^{2}}$ C. $f=0,g=-2a,c=3{{a}^{2}}$ D. none of these

Last updated date: 19th Jul 2024
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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.

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 ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ .