The circles x^2 + y^2 + 2x -2y + 1 = 0 and x^ | vedclass

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Question

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} \

Vedclass · Accepted Answer

externally at $(0, 1)$

Vedclass · Answer

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

$(0,1)$

The circles $x^2 + y^2 + 2x -2y + 1 = 0$ and $x^2 + y^2 -2x -2y + 1 = 0$ touch each other :-

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