AP EAMCET Engineering 2011 Solved Paper

Given equation of circles are
S1≡x2+y2+2x+2y+1=0
and S2≡x2+y2−2x+2y+1=0
In circle S1, centre c1=(−1,−1), radius
r1=√1+1−1=1
In circle S2, centre C2=(1,−1), radius
r2=√1+1−1=1
Now, distance between two centre C1C2=√(−1−1)2+0
=2
Here, we see that
C1C2=r1+r2
i.e., both circle touch externally.
So, point of contact of circle S1 and S2
= Mid-point of C1 and C2 =(

−1+1

2

,

−1−1

2

)=(0,−1)