For how many values of $p$ the circle $x^{2} + y^{2} + 2 x + 4 y - p = 0$ and the co-ordinate axes have exactly three common points?

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Solution

$x^{2} + y^{2} + 2 x + 4 y - p = 0 C a s e (I) {(0)}^{2} + {(0)}^{2} + 0 + 0 p = 0 C a s e (I I) g^{2} - c = 0 f^{2} - c > 0 1 - (- p) = 0 p = - 1 4 - (- p) > 0 p > - 4 C a s e (I I I) f^{2} - c = 0 4 - (p) = 0 p = - 4 g^{2} - c > 0 1 - (- p) > 0 p > - 1$
$∴$ Total 2 values of p, the circle and the co-ordinate have 3 common point.

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