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The circles $x^2 + y^2 + 2x -2y + 1 = 0$ and $x^2 + y^2 -2x -2y + 1 = 0$ touch each other :-
externally at $(0, 1)$
internally at $(0, 1)$
externally at $(1, 0)$
internally at $(1, 0)$
Solution
The centres of the two circles are $\mathrm{C}_{1}(-1,1)$ and
$\mathrm{C}_{2}(1,1)$ and both have radii equal to $1 .$ We have
$\mathrm{C}_{1} \mathrm{C}_{2}=2$ and sum of the radii $=2$
So, the two circles touch each other externally.
The equation of the common tangent is obtained by subtracting the two equations.
The equation of the common tangnet is
$4 \mathrm{x}=0 \Rightarrow \mathrm{x}=0$
Putting $\mathrm{x}=0$ in the equation of the either circle,
we get
$\mathrm{y}^{2}-2 \mathrm{y}+1=0 \Rightarrow(\mathrm{y}-1)^{2}=0 \mathrm{y}=1$
Hence, the points where the two circles touch is
$(0,1)$
