Question 14354
The equation of a circle with centre as (h,k) and radius r is given by the formula
  {{{(x-h)^2+(y-k)^2=r^2}}}.
  So we have to write the given equation in this form to get the centre of the circle.
  {{{x^2+y^2-4*x+6*y+4=0}}} OR {{{x^2-4*x+y^2+6*y+4=0}}}  
 {{{(x-2)^2-4+(y+3)^2-9+4=0}}}...Note that we have added and subtracted 4 and 9 to make up the squares mentioned in brackets.
 Hence {{{(x-2)^2+(y+3)^2=9}}}
  Hence (2,-3)is the centre of this circle .Similarly the centre of the second circle is obtained from 
   {{{x^2+y^2+6*x+4*y+9=0}}} OR {{{x^2+6*x+y^2+4*y+9=0}}}  
 {{{(x+3)^2-9+(y+2)^2-4+9=0}}}...Note that we have added and subtracted 9 and 4 to make up the squares mentioned in brackets.
 Hence {{{(x+3)^2+(y+2)^2=4}}}
  Hence the centre of the second circle is (-3,-2)
  The equation of line joining 2 ponts  (x1,y1) and (x2,y2)is given by the formula 
 {{{Y-y1=(y2-y1)*(X-x1)/((x2-x1))}}}
 Substituting (2,-3) and (-3,-2) in the above formula we get 
 Y-(-3)=(-2-(-3))*(X-2)/(-3-(-2))
  Y+3=(X-2)/(-5)
-5Y-15=X-2
X+5Y+13=0