The given equation of the circle is:
$x^2 + y^2 + 2x + 4y - p = 0$ We need to determine for how many values of $p$ the circle and the coordinate axes intersect at exactly three points. Let’s analyze step by step how the circle intersects the $x$-axis and $y$-axis.

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1. Intersections with the coordinate axes

Intersection with the x-axis ($y = 0$)

By substituting $y = 0$ into the equation: $x^2 + 2x - p = 0$ This is a quadratic equation in $x$. The roots of this equation are: $x = \frac{-2 \pm \sqrt{4 + 4p}}{2} = -1 \pm \sqrt{1 + p}$ For real solutions, $1 + p \geq 0 \implies p \geq -1$.

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Intersection with the y-axis ($x = 0$)

By substituting $x = 0$ into the equation: $y^2 + 4y - p = 0$ This is a quadratic equation in $y$. The roots of this equation are: $y = \frac{-4 \pm \sqrt{16 + 4p}}{2} = -2 \pm \sqrt{4 + p}$ For real solutions, $4 + p \geq 0 \implies p \geq -4$.

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2. Exactly three points of intersection

We are asked to determine the value of $p$ such that the circle intersects the coordinate axes at exactly three points. This can happen in two possible ways:

1. Two points on one axis and one point on the other.
2. Since the roots are symmetric (two solutions per axis), the only way to have exactly three points is if one of the quadratic equations gives repeated roots (a double root) and the other gives two distinct real roots.

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3. Identifying when one axis has a double root

Case 1: Double root on the x-axis

A double root occurs if the discriminant of the equation for the x-axis is zero: $(1 + p) = 0 \implies p = -1$ For $p = -1$, the roots on the x-axis are: $x = -1 \quad (\text{repeated root}).$ Thus, the circle touches the x-axis at $(-1, 0)$.

Now, substitute $p = -1$ in the equation for the y-axis: $y^2 + 4y + 1 = 0.$ The discriminant is: $16 - 4 = 12 > 0,$ which means there are two distinct points of intersection on the y-axis. So, for $p = -1$, the circle intersects the axes at exactly three points: one on the x-axis and two on the y-axis.

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Case 2: Double root on the y-axis

A double root occurs if the discriminant of the equation for the y-axis is zero: $4 + p = 0 \implies p = -4.$ For $p = -4$, the roots on the y-axis are: $y = -2 \quad (\text{repeated root}).$ Thus, the circle touches the y-axis at $(0, -2)$.

Now, substitute $p = -4$ in the equation for the x-axis: $x^2 + 2x + 4 = 0.$ The discriminant is: $4 - 16 = -12 < 0,$ which means there are no real points of intersection on the x-axis. So, for $p = -4$, the circle intersects the y-axis at only one point, not three.

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4. Conclusion

The only valid value of $p$ that results in exactly three points of intersection is: $p = -1.$

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Final Answer

The circle and the coordinate axes have exactly three common points for one value of $p$, which is: $\boxed{p = -1}.$

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Do you have any questions, or would you like more details?

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Related Questions

1. What happens when $p = 0$?
2. How do we calculate the center and radius of this circle?
3. How many intersection points occur if $p > 0$?
4. What are tangency conditions between a circle and a line?
5. Can a circle intersect the axes at more than four points?

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Tip

For a quadratic equation $ax^2 + bx + c = 0$ to have a double root, its discriminant $\Delta = b^2 - 4ac$ must be exactly zero.