Question 971063
Since centres of circles lie on y+3=0, i.e y= -3
Let coordinates of centres be (x,-3)
Since the circles touch the line -2x+3y=0
So, the distance from centre of circles to this line are radii of circles.
So, Using the distance formula i.e 

(|-2x+3*(-3)|)/(sqrt((-2^2)+3^2)) = sqrt(13)  (|..| = mod sign)
or, |-2x+3*(-3)| = 13
or, -2x-9 = (+-13)
So, {{{-2x-9 = 13}}} -(i) or {{{-2x-9 = -13}}}- (ii)
Solving both eqns. separately,
we get, {{{x= -11}}} or {{{x=2}}}
So, the coordinates of center are (-11,-3) and (2,-3)
Equations of circles are {{{(x+11)^2 + (y+3)^2 = 13}}} and {{{(x-2)^2 + (y+3)^2 = 13}}}
i.e. {{{x^2 + y^2 + 22x + 6y -117 =0}}} and {{{x^2 + y^2 - 4x + 6y =0}}}