Hint: We first try to form the given circle in its general form of (x−α)2+(y−β)2=r2 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 x2+y2+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 x2+y2+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 x2+y2+2gx+2fy+c=0 . We transform it in its general form of (x−α)2+(y−β)2=r2 and get (x−g)2+(y−f)2=(g2+f2−c−−−−−−−−−√)2 . O is the centre. Equating with the general equation of circle (x−α)2+(y−β)2=r2 , we get the centre as O≡(−g,−f) and the radius as g2+f2−c−−−−−−−−−√ units. We can put the value −f=0 which means f=0 . Also, the radius is a. This means the centre is O≡(−g,0) and the radius as g2+f2−c−−−−−−−−−√=a units. Solving the radius, we get g2−c=a2 . Now we put the value of y=0 in the equation x2+y2+2gx+2fy+c=0 to get the two end points of the diameter. This gives x2+2gx+c=0 . Solving we get x=−2g±4g2−4c−−−−−−−√2=−g±g2−c−−−−−√ . Now we place the given two options which are possible and they are f=0,g=a,c=3a2 and f=0,g=−2a,c=3a2 . In both cases f=0,c=3a2 . Now we put c=3a2 in g2−c=a2 . We get g2=4a2 which gives g=±2a . Therefore, the correct option is C. Note: We need to remember that the general equation x2+y2+2gx+2fy+c=0 always makes an intercept 2g2−c−−−−−√ units on the X-axis and 2f2−c−−−−−√ units on the Y-axis. We use that to find the diameter length in the circle of x2+y2+2gx+2fy+c=0 .