SOLUTION: 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.

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.
Answer by dkppathak(439) (Show Source): You can put this solution on YOUR website!
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)

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