Number of Real Parameter Values for Circle Intersecting Coordinate Axes

To find the total count of real values of $p$ for which the circle $x^2 + y^2 + 2x + 4y - p = 0$ has precisely three distinct intersection points with the coordinate axes, we analyze the intersection points with each axis.

1. Intersections with the $x$-axis

Setting $y = 0$ in the circle's equation yields: $x^2 + 2x - p = 0$ The discriminant of this quadratic equation is: $\Delta_x = 2^2 - 4(1)(-p) = 4(1 + p)$

• If $p > -1$, $\Delta_x > 0 \implies$ there are $2$ distinct intersection points on the $x$-axis.
• If $p = -1$, $\Delta_x = 0 \implies$ there is $1$ intersection point on the $x$-axis.
• If $p < -1$, $\Delta_x < 0 \implies$ there are $0$ intersection points on the $x$-axis.

2. Intersections with the $y$-axis

Setting $x = 0$ in the circle's equation yields: $y^2 + 4y - p = 0$ The discriminant of this quadratic equation is: $\Delta_y = 4^2 - 4(1)(-p) = 4(4 + p)$

• If $p > -4$, $\Delta_y > 0 \implies$ there are $2$ distinct intersection points on the $y$-axis.
• If $p = -4$, $\Delta_y = 0 \implies$ there is $1$ intersection point on the $y$-axis.
• If $p < -4$, $\Delta_y < 0 \implies$ there are $0$ intersection points on the $y$-axis.

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3. Finding values of $p$ for exactly 3 distinct intersection points

We divide the problem into two main cases based on whether the circle passes through the origin $(0,0)$:

Case 1: The circle passes through the origin $(0,0)$

Substituting $(0,0)$ into the circle equation gives: $0^2 + 0^2 + 2(0) + 4(0) - p = 0 \implies p = 0$

For $p = 0$:

• Intersections with the $x$-axis: $x^2 + 2x = 0 \implies x = 0$ and $x = -2$, giving points $(0,0)$ and $(-2,0)$.
• Intersections with the $y$-axis: $y^2 + 4y = 0 \implies y = 0$ and $y = -4$, giving points $(0,0)$ and $(0,-4)$.

The set of all intersection points with the axes is $\{ (0,0), (-2,0), (0,-4) \}$, which contains exactly $3$ distinct points. Thus, $p = 0$ is a valid value.

Case 2: The circle does not pass through the origin $(0,0)$

Here, $p \neq 0$, meaning $(0,0)$ is not an intersection point. Consequently, the intersection points on the $x$-axis and the $y$-axis are completely disjoint.

The total number of distinct intersection points is given by $N = N_x + N_y$, where $N_x$ and $N_y$ are the number of intersections on the $x$-axis and $y$-axis, respectively. We require $N = 3$:

1. Subcase 2a: $N_x = 1$ and $N_y = 2$

   • $N_x = 1 \implies p = -1$.
   • Substituting $p = -1$ into $N_y$: since $p = -1 > -4$, we have $\Delta_y = 4(4 - 1) = 12 > 0 \implies N_y = 2$.
   • The total number of points is $N = 1 + 2 = 3$.
   • Thus, $p = -1$ is a valid value.

2. Subcase 2b: $N_x = 2$ and $N_y = 1$

   • $N_y = 1 \implies p = -4$.
   • Substituting $p = -4$ into $N_x$: since $p = -4 < -1$, we have $\Delta_x = 4(1 - 4) = -12 < 0 \implies N_x = 0$.
   • The total number of points is $N = 0 + 1 = 1 \neq 3$.
   • Thus, $p = -4$ does not yield $3$ intersection points.

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Conclusion

The only real values of $p$ satisfying the condition are $p = 0$ and $p = -1$.

Therefore, the total count of real values of $p$ is 2.