The number of common tangents to the circles $x^{2} + y^{2} + 2 x + 8 y - 23 = 0 and x^{2} + y^{2} - 4 x - 10 y + 9 = 0$ are .

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Solution

The correct option is D 2
For circle $x^{2} + y^{2} + 2 x + 8 y - 23 = 0$ , let A be the centre and $r_{1}$ be the radius.
Hence, the coordinates of A are (-1,-4) and $r_{1}$ = $\sqrt{40}$
For circle $x^{2} + y^{2} - 4 x - 10 y + 9 = 0$ , let B be the centre and $r_{2}$ be the radius.
Hence, the coordinates of B are (2,5) and $r_{2}$ = $\sqrt{20}$ .
Also, AB= $\sqrt{90}$ .
Clearly, $r_{1}$ + $r_{2}$ >AB
$⟹$ The two circles intersect each other,
Hence, the number of common tangents = 2

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