The number of common tangents of the circles ${(x + 2)}^{2} + {(y - 2)}^{2} = 49$ and ${(x - 2)}^{2} + {(y + 1)}^{2} = 4$ are

$0$

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$1$

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$3$

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$4$

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Solution

The correct option is B
$1$

Explanation for correct option
The given circle have centre at $A (- 2, 2), B (2, - 1)$
$A B = \sqrt{{(2 + 2)}^{2} + {(- 1 - 2)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 = r_{1} - r_{2}$
Where $r_{1}$ is the radius of the circle with center at $A$ and $r_{2}$ is the radius of circle with center at $B$
Therefore, the given circle touch internally
Therefore there is only one common tangent
Hence, option(B) is correct

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