Solving $$y = mx$$   and $${x^2} + {y^2} - 2x - 2y = 0,$$     we get
$$\eqalign{ & {x^2} + {m^2}{x^2} - 2x - 2mx = 0 \cr & \Rightarrow \,x = 0,\,\frac{{2\left( {1 + m} \right)}}{{1 + {m^2}}} \cr} $$
Similarly, solving $$y = - mx$$   and the equation of the circle, we get
$$\eqalign{ & x = 0,\,\frac{{2\left( {1 - m} \right)}}{{1 + {m^2}}} \cr & \therefore \,A = \left( {\frac{{2\left( {1 + m} \right)}}{{1 + {m^2}}},\,\frac{{2m\left( {1 + m} \right)}}{{1 + {m^2}}}} \right)\,{\text{and }}B = \left( {\frac{{2\left( {1 - m} \right)}}{{1 + {m^2}}},\,\frac{{ - 2m\left( {1 - m} \right)}}{{1 + {m^2}}}} \right) \cr} $$
Let the middle point of $$AB$$  be $$\left( {\alpha ,\,\beta } \right).$$  Then
$$\eqalign{ & \alpha = \frac{1}{2}\left( {\frac{{2\left( {1 + m} \right)}}{{1 + {m^2}}} + \frac{{2\left( {1 - m} \right)}}{{1 + {m^2}}}} \right){\text{ and }}\beta = \frac{1}{2}\left( {\frac{{2m\left( {1 + m} \right)}}{{1 + {m^2}}} + \frac{{ - 2m\left( {1 - m} \right)}}{{1 + {m^2}}}} \right) \cr & \therefore \,\,\alpha = \frac{2}{{1 + {m^2}}},\,\beta = \frac{{2{m^2}}}{{1 + {m^2}}} \cr} $$
Eliminating m from these, $$\alpha + \beta = 2.$$

The given circle is $${x^2} + {y^2} + 2x + 4y - 3 = 0$$

Centre $$\left( { - 1,\, - 2} \right)$$
Let $$Q\left( {\alpha ,\,\beta } \right)$$  be the point diametrically opposite to the point $$P\left( {1,\,0} \right),$$
then $$\frac{{1 + \alpha }}{2} = - 1{\text{ and }}\frac{{0 + \beta }}{2} = - 2$$
$$ \Rightarrow \alpha = - 3,\,\,\beta = - 4,\,\,\,{\text{So }}Q\,\,{\text{is}}\left( { - 3,\, - 4} \right)$$

Coordinates of the centre of given circle $$ = \left( {\alpha ,\,\alpha } \right)$$
and radius $$ = \sqrt {{{\left( \alpha \right)}^2} + {{\left( \alpha \right)}^2} - {\alpha ^2}} = \sqrt {{\alpha ^2}} = \alpha $$

$$\eqalign{ & \therefore \,{\left( {\alpha - 5} \right)^2} + {\left( \alpha \right)^2} = {\left( \alpha \right)^2} \cr & \Rightarrow {\alpha ^2} + 25 - 10\alpha = 0 \cr & \Rightarrow {\left( {\alpha - 5} \right)^2} = 0 \cr & \Rightarrow \alpha = 5 \cr} $$
Then, other root will always real.

$${x^2} + {y^2} - 4x - 6y - 12 = 0.....({\text{i}})$$
Centre, $${c_1} = \left( {2,\,3} \right)$$    and Radius, $${r_1} = 5\,{\text{units}}$$
$${x^2} + {y^2} + 6x + 18y + 26 = 0.....({\text{ii}})$$
Centre, $${c_2} = \left( { - 3,\, - 9} \right)$$    and Radius, $${r_2} = 8\,{\text{units}}$$
$$\eqalign{ & {C_1}{C_2} = \sqrt {{{\left( {2 + 3} \right)}^2} + {{\left( {3 + 9} \right)}^2}} = 13\,{\text{units}} \cr & {r_1} + {r_2} = 5 + 8 = 13 \cr & \therefore {C_1}{C_2} = {r_1} + {r_2} \cr} $$

Therefore there are three common tangents.

$$\eqalign{ & {\text{Here, }}{\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} = 0 \cr & \Rightarrow x = 1,\,y = - 2,\,{\text{which give a point}}{\text{.}} \cr} $$

The locus is the radical axis which is perpendicular to the line joining the centres of the circles.

$$\eqalign{ & f\left( {x,\,y} \right) \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0 \cr & {\text{Now,}}\,{\text{ }}f\left( {0,\,\lambda } \right) \equiv {\lambda ^2} + 2f\lambda + c = 0,\,{\text{and its roots are }}2,\,2 \cr & \therefore \,\,2 + 2 = - 2f,\,2 \times 2 = c,\,\,{\text{i}}{\text{.e}}{\text{.,}}f = - 2,\,c = 4 \cr & f\left( {\lambda ,\,0} \right) \equiv {\lambda ^2} + 2g\lambda + c = 0,\,{\text{and its roots are }}\frac{4}{5},\,5 \cr & \therefore \,\,\frac{4}{5} + 5 = - 2g,\,\frac{4}{5} \times 5 = c \cr & {\text{Thus }}g = - \frac{{29}}{{10}},\,f = - 2 \cr} $$

The circle is $${\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} = {3^2}.$$      As any tangent to $${x^2} + {y^2} = {3^2}$$   is given by $$y = mx + 3\sqrt {1 + {m^2}} ,$$     any tangent to the given circle will be $$y + 2 = m\left( {x - 1} \right) + 3\sqrt {1 + {m^2}} .$$

$$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$$       (formula for orthogonal intersection of two cricles)
$$\eqalign{ & \Rightarrow 2(1)(0) + 2(k)(k) = 6 + k \cr & \Rightarrow 2{k^2} - k - 6 = 0 \cr & \Rightarrow k = - \frac{3}{2},\,\,2\, \cr} $$

Let the circle be $${x^2} + {y^2} + 2gx + 2fy = 0.$$     It cuts the other circle orthogonally if $$2g\left( { - \frac{5}{2}} \right) + 2f.\frac{3}{2} = 0 + \left( { - 1} \right),$$       i.e., $$5\left( { - g} \right) - 3\left( { - f} \right) + 1 = 0.$$
Hence, the locus of the centre $$\left( { - g,\, - f} \right)$$   is $$5x - 3y + 1 = 0.$$