For how many value of p, the circle x2 + y2 + 2x + 4y - p = 0 and the coordinate axes have exactly three common points

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.

---

### 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, \(1 + p \geq 0 \implies p \geq -1\).

---

#### **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, \(4 + p \geq 0 \implies p \geq -4\).

---

### 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.

---

### 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.

---

#### 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.

---

### 4. Conclusion

The only valid value of \(p\) that results in exactly three points of intersection is:
\[
p = -1.
\]

---

### Final Answer

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

---

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?

---

### 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.

For how many values of p, the circle x^2 + y^2 + 2x + 4y - p = 0 and the coordinate axes have exactly three common points?

Discover how to determine the value of p for which the circle x^2 + y^2 + 2x + 4y - p = 0 intersects the coordinate axes at exactly three points. This solution explores quadratic equations and the discriminant, suitable for students in Grades 10-12.

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