(c) Given, equations of the circles

${x^2} + {y^2} – 2x + 6y + 6$$=0$ …..$(i)$

and ${x^2} + {y^2} – 5x + 6y + 15 = 0$ …..$(ii)$

We know that the standard equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0.$

Therefore for circle $(i),$ $g = – 1;\,f = 3;\,\,\,c = 6;$ centre $A = (1,\, – 3)$

and radius $({r_1}) = \sqrt {{g^2} + {f^2} – c} = \sqrt {1 + 9 – 6} = 2$.

Similarly, for circle $(ii),$ $g = \frac{{ – 5}}{2};\,\,f = 3;\,c = 15;$

 Centre $B \equiv \,\left( { + \frac{5}{2}, – 3} \right)$

and radius $({r_2}) = \sqrt {\frac{{25}}{4} + 9 – 15} = \frac{1}{2}$

Therefore distance between $A$ and $B$ $ = \sqrt {{{\left( {\frac{5}{2} – 1} \right)}^2} + {{( – 3 + 3)}^2}} = \frac{3}{2}$

and difference of radii $({r_1} – {r_2}) = 2 – \frac{1}{2} = \frac{3}{2}.$

Since distance between $A$ and $B$ is equal to ${r_1} – {r_2},$

therefore the circles touch each other internally.