Center of $$S$$ : $$O(-3, 2)$$ center of given circle $$A(2, -3)$$

$$ \Rightarrow OA = 5\sqrt 2 $$

Also $$AB=5$$ (as $$AB=r$$ of the given circle)

$$ \Rightarrow $$ Using pythagoras theorem in $$\Delta OAB$$

$$r = 5\sqrt 3 $$

For the given circle,

center : $$(4,4)$$

radius $$=6$$

$$6 + k = \sqrt {{{\left( {h - 4} \right)}^2} + {{\left( {k - 4} \right)}^2}} $$

$${\left( {h - 4} \right)^2} = 20k + 20$$

$$\therefore$$ locus of $$(h, k)$$ is

$${\left( {h - 4} \right)^2} = 20\left( {y + 1} \right),$$

which is parabola.

$${x^2} + {y^2} - 4x - 6y - 12 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Center, $${c_1} = \left( {2,\,3} \right)$$ and Radius, $${r_1} = 5$$ units

$${x^2} + {y^2} + 6x + 18y + 26 = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Center, $${c_2} = \left( { - 3, - 9} \right)$$ and Radius, $${r_2} = 8$$ units

$${C_1}{C_2} = \sqrt {{{\left( {2 + 3} \right)}^2} + {{\left( {3 + 9} \right)}^2}} = 13\,\,$$ units

$${r_1} + {r_2} = 5 + 8 = 13$$

$$\therefore$$ $${C_1}{C_2} = {r_1} + {r_2}$$



Therefore there are three common tangents.

Intersection point of $$2x - 3y + 4 = 0$$

and $$x-2y+3=0$$ is $$(1, 2)$$



Since, $$P$$ is the fixed point for given family of lines

So, $$PB=PA$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2$$

$${\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt 2 } \right)^2}$$

$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$$

Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\sqrt 2 .$$