For a circle to meet $$x$$-axis in two points on the opposite side of the origin its radius $$r,$$ should be more the distance of its centre from the origin.
Co-ordinate of centre of the circle $${x^2} + {y^2} + 2gx + 2fy + c = 0$$       is $$\left( { - g,\, - f} \right)\,:$$

In the figure shown,
$$OQ = OP = r,$$    and distance of centre $$C,$$  from origin, $$O$$ is $$CO$$
$$\eqalign{ & r > \sqrt {OC} \,{\text{ i}}{\text{.e}}{\text{., }}r > \sqrt {{{\left( { - g} \right)}^2} + {{\left( { - f} \right)}^2}} \cr & {\text{or, }}\sqrt {{{\left( { - g} \right)}^2} + {{\left( { - f} \right)}^2} - c} > \sqrt {{{\left( { - g} \right)}^2} + {{\left( { - f} \right)}^2}} \cr & {\text{or, }}{g^2} + {f^2} - c > {g^2} + {f^2} \cr & {\text{or, }} - c > 0 \cr & {\text{or, }}c < 0 \cr} $$

The given circles are concentric with centre at $$\left( {0,\,0} \right)$$  and the length of the perpendicular from $$\left( {0,\,0} \right)$$  on the given line is $$p$$.

Let $$OL = p$$
then, $$AL = \sqrt {O{A^2} - O{L^2}} = \sqrt {{a^2} - {p^2}} $$
and $$PL = \sqrt {O{P^2} - O{L^2}} = \sqrt {{b^2} - {p^2}} $$
$$ \Rightarrow AP = \sqrt {{a^2} - {p^2}} - \sqrt {{b^2} - {p^2}} $$

Let $$C\left( {h,\,k} \right)$$   be the centre of circle touching $${x^2} = y$$  at $$B\left( {2,\,4} \right).$$  Then equation of common tangent at $$B$$ is

$$2.x = \frac{1}{2}\left( {y + 4} \right)\,\,i.e.,\,\,\,4x - y = 4$$
Radius is perpendicular to this tangent
$$\eqalign{ & \therefore 4\left( {\frac{{k - 4}}{{h - 2}}} \right) = - 1\,\, \Rightarrow 4k = 18.....(1) \cr & {\text{Also }}\,AC = BC \cr & \Rightarrow {h^2} + {\left( {k - 1} \right)^2} = {\left( {h - 2} \right)^2} + {\left( {k - 4} \right)^2} \cr & \Rightarrow 4h + 6k = 19.....(2) \cr} $$
Solving (1) and (2) we get the centre as $$\left( {\frac{{ - 16}}{5},\,\frac{{53}}{{10}}} \right)$$

Equation of pair tangents to the circle $${x^2} + {y^2} + 4x + 2y - 4 = 0$$      from the point $$\left( {1,\,\frac{1}{2}} \right)$$  is
$$\eqalign{ & S{S_1} = {T^2}\left( {{x^2} + {y^2} + 4x + 2y - 4} \right)\left( {1 + \frac{1}{4} + 4 + 1 - 4} \right) \cr & = {\left( {x + \frac{1}{2}y + 2\left( {x + 1} \right)y + \frac{1}{2} - 4} \right)^2} \cr & \Rightarrow \frac{9}{4}\left( {{x^2} + {y^2} + 4x + 2y - 4} \right) = {\left( {3x + \frac{3}{2}y - \frac{3}{2}} \right)^2} \cr & \Rightarrow \frac{9}{4}{x^2} + \frac{9}{4}{y^2} + 9x + \frac{9}{2}y - 9 = 9{x^2} + \frac{9}{4}{y^2} + \frac{9}{4} + 9xy - \frac{9}{2}y - 9x \cr & \Rightarrow \frac{{27}}{4}{x^2} - 18x - 9y + 9xy + \frac{{45}}{4} = 0 \cr & \Rightarrow \frac{3}{4}{x^2} + xy - 2x - y + \frac{5}{4} = 0 \cr & \Rightarrow 3{x^2} + 4xy - 8x - 4y + 5 = 0 \cr} $$
It $$\theta $$ is the angle of these lines, then
$$\eqalign{ & \tan \,\theta = \frac{{2\sqrt {{h^2} - ab} }}{{a + b}} \cr & \Rightarrow \tan \,\theta = \frac{{2\sqrt 4 }}{3} \cr & \Rightarrow \tan \,\theta = \frac{4}{3} \cr & \Rightarrow \sin \,\theta = \frac{4}{5} \cr & \Rightarrow \theta = {\sin ^{ - 1}}\left[ {\frac{4}{5}} \right] \cr} $$

Let the variable circle be
$$\eqalign{ & {x^2} + {y^2} + 2gx + 2fy + c = 0.....(1) \cr & \therefore {p^2} + {q^2} + 2gp + 2fq + c = 0.....(2) \cr} $$
Circle (1) touches $$x$$-axis,
$$\eqalign{ & \therefore {g^2} - c = 0 \Rightarrow c = {g^2}\,.\,{\text{From }}(2) \cr & {p^2} + {q^2} + 2gp + 2fq + {g^2} = 0.....(3) \cr} $$
Let the other end of diameter through $$\left( {p,\,q} \right)$$  be $$\left( {h,\,k} \right)$$
then, $$\frac{{h + p}}{2} = - g$$   and $$\frac{{k + q}}{2} = - f$$
Put in (3)
$$\eqalign{ & {p^2} + {q^2} + 2p\left( { - \frac{{h + p}}{2}} \right) + 2q\left( { - \frac{{k + q}}{2}} \right) + {\left( {\frac{{h + p}}{2}} \right)^2} = 0 \cr & \Rightarrow {h^2} + {p^2} - 2hp - 4kq = 0 \cr} $$
$$\therefore $$ locus of $$\left( {h,\,k} \right)$$  is $${x^2} + {p^2} - 2xp - 4yq = 0$$
$$ \Rightarrow {\left( {x - p} \right)^2} = 4qy$$

Line $$5x - 2y + 6 = 0$$    is intersected by tangent at $$P$$ to circle $${x^2} + {y^2} + 6x + 6y - 2 = 0$$      on $$y$$-axis at $$Q\,\left( {0,\,3} \right).$$
In other words tangent passes through $$\left( {0,\,3} \right)$$
$$\therefore \,\,PQ = $$     length of tangent to circle from $$\left( {0,\,3} \right)$$
$$\eqalign{ & = \sqrt {0 + 9 + 0 + 18 - 2} \cr & = \sqrt {25} \cr & = 5 \cr} $$

In an equilateral triangle, the circumcentre and the incentre are the same point.
$$\therefore $$  incentre $$ = \left( { - g,\, - f} \right)$$
Also $${1^2} + {1^2} + 2g + 2f + c = 0$$       (given), i.e., $$c = - 2\left( {g + f + 1} \right)$$
Also, in an equilateral triangle, circumradius $$2 \times $$  inradius
$$\therefore $$  inradius $$ = \frac{1}{2} \times \sqrt {{g^2} + {f^2} - c} $$
$$\therefore $$  the equation of the incircle is
$$\eqalign{ & {\left( {x + g} \right)^2} + {\left( {y + f} \right)^2} \cr & = \frac{1}{4}\left( {{g^2} + {f^2} - c} \right) \cr & = \frac{1}{4}\left( {{g^2} + {f^2}} \right) + \frac{1}{4}.2\left( {g + f + 1} \right){\text{ Simplify}}{\text{.}} \cr} $$

The line joining $$\left( {4,\,3} \right)$$  and $$\left( {2,\,1} \right)$$  is also along a diameter. So, the centre is the intersection of the diameters $$2x - y = 2$$   and $$y - 3 = \frac{{3 - 1}}{{4 - 2}} = \left( {x - 4} \right).$$
Solving these, the centre $$ = \left( {1,\,0} \right)$$
Also, the radius $$=$$ the distance between $$\left( {1,\,0} \right)$$  and $$\left( {2,\,1} \right) = \sqrt 2 .$$

Let $$A,\,B,$$  be the centers of the two circles,
Slope of the common tangent $$ = - \frac{{12}}{5}$$

$$\therefore $$  Slope of $$AB$$  is $$\tan \,\theta = - \frac{1}{{ - \frac{{12}}{5}}} = \frac{5}{{12}}$$
The point $$\left( {1,\, - 1} \right)$$  lies on the line $$AB$$  and the points $$A$$ and $$B$$ are at a distance $$13$$  from the point $$\left( {1,\, - 1} \right)$$
$$\therefore $$  Coordinates of $$A$$ and $$B$$ are $$\left( {1 \pm 13\,\cos \,\theta ,\, - 1 \pm 13\,\sin \,\theta } \right),$$      where $$\tan \,\theta = \frac{5}{{12}}$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.,}}\,\,\left( {1 \pm 13\frac{{12}}{{13}},\, - 1 \pm 13\frac{5}{{13}}} \right){\text{ or }}\left( {1 \pm 12,\, - 1 \pm 5} \right) \cr & {\text{i}}{\text{.e}}{\text{.,}}\,\,\left( {13,\,4} \right){\text{ and }}\left( { - 11,\, - 6} \right) \cr} $$

Centres and radii of two circles are $${C_1}\left( {5,\,0} \right);\,3 = {r_1},$$     and $${C_2}\left( {0,\,0} \right);\,r = {r_2}$$
As circles intersect each other in two distinct points,
$$\eqalign{ & \left| {{r_1} - {r_2}} \right| < {C_1}{C_2} < {r_1} + {r_2} \cr & \Rightarrow \left| {r - 3} \right| < 5 < r + 3 \cr & \Rightarrow 2 < r < 8 \cr} $$