The point of contact of the circles $x^2+y^2+2 x+2 y+1=0$ and $\quad x^2+y^2-2 x+2 y+1=0$ is
The point of contact of the circles
$x^2+y^2+2 x+2 y+1=0$
and $\quad x^2+y^2-2 x+2 y+1=0$ is
(0, 1)
(0,-1)
(1, 0)
(-1, 0)
Solution
Given equation of circles are
$\begin{aligned}
S_1 & \equiv x^2+y^2+2 x+2 y+1=0 \\
\text { and } \quad S_2 \equiv x^2+y^2-2 x+2 y+1 & =0
\end{aligned}$
In circle $S_1$, centre $c_1=(-1,-1)$, radius $r_1=\sqrt{1+1-1}=1$
In circle $S_2$, centre $C_2=(1,-1)$, radius $r_2=\sqrt{1+1-1}=1$
Now, distance between two centre
$\begin{gathered}
C_1 C_2=\sqrt{(-1-1)^2+0} \\
=2
\end{gathered}$
Here, we see that
$C_1 C_2=r_1+r_2$
i.e., both circle touch externally.
So, point of contact of circle $S_1$ and $S_2$
$=$ Mid-point of $C_1$ and $C_2$
$=\left(\frac{-1+1}{2}, \frac{-1-1}{2}\right)=(0,-1)$