Question 989022
{{{x^2 + y^2- 2x-6y+ 9= 0}}}
=> {{{(x-1)^2 + (y-3)^2 = 1}}}
So, centre of given circle, is (1,3) and radius 1.

Let centre of required circles be (a,0), radius given = 4
Draw figure joining centres and applying pythagoras theorem, we get,

{{{(1-a)^2 + 3^2 = (r+R)^2}}}
=> {{{a^2-2a+1+9=(1+4)^2}}}
=> {{{a^2-2a-15 = 0}}}
=> a = 5 or a = -3

So, centre of required circles are (5,0) and (-3,0)
And equations are:-
{{{(x-5)^2 + y^2 = 4^2}}} and {{{(x+3)^2 + y^2 = 4^2}}}
=> {{{x^2+y^2-10x+9=0}}} and {{{x^2+y^2+6x-7=0}}}