Question 1168445
The circles (x+1)^2+(y-1)^2=9 and (x-3)^2+(y+1)^2=9 are overlapped. Find the point of intersections of the two circles.
solution
x^2+1+2x+y^2+1-2y=9    and  x^2+9-6x+y^2+1+2y=9  both are equal to 9 by equalizing both circle 
x^2+1+2x+y^2+1-2y=  x^2+9-6x+y^2+1+2y
2x-2y+2=-6x+2y+10
x-y+1=-3x+y+5 
4x=2y+4
2x=y+2   (1)   y=2x-2 putting value in one equation of circle 
(x+1)^2+(2x-3)^2=9
x^2+1+2x+4x^2+9-12x=9    
5x^2-10x+10=9    
5x^2-10x+1=0    
               centers are   ( -1,1  )  and (3,-1  )
 distance 2 sqre root 5  distance between both radius is less than  sum of radius therefore both circle are intersecting
x=10+4sqr root5/10 or 10-4sqr root 5/10
x=1.89   or 0.105 

y=1.78  or   y=-1.79
point of intersection will be (1.89,1.78)  and (0.105,-1.79)