The shortest distance between the circles $(x - 1)^{2} + (y + 2)^{2} = 1$ and $(x + 2)^{2} + (y - 2)^{2} = 4$ is

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Solution

The correct option is B 2
Equation of first circle is $(x - 1)^{2} + (y + 2)^{2} = 1$ .
Hence, centre of the circle is $(1, - 2)$ and radius of the circle is $1$
Equation of second circle is $(x + 2)^{2} + (y - 2)^{2} = 4$ .
Hence, centre of the circle is $(- 2, 2)$ and radius of the circle is $2$
Shortest distance between the two circles $= A B = C_{1} C_{2} - A C_{1} - B C_{2}$
$= \sqrt{(- 2 - 1)^{2} + (2 - (- 2))^{2}} - 1 - 2$
$= 5 - 3 = 2$

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