Animated Solution for Mathematics - Circles: 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 ?

The circle is defined by the equation:

By comparing this to the general form $x^2+y^2+2gx+2fy+c=0$, we identify the center of the circle at $(-1,-2)$. The circle's interaction with the coordinate axes depends entirely on the parameter $p$.

Consider the case where the circle passes through the origin $(0,0)$. Substituting $x=0$ and $y=0$ into the equation yields:

When $p=0$, the equation becomes $x^2+y^2+2x+4y=0$. For the $x$-axis ($y=0$), we get $x^2+2x=0$, resulting in points $(0,0)$ and $(-2,0)$. For the $y$-axis ($x=0$), we get $y^2+4y=0$, resulting in points $(0,0)$ and $(0,-4)$.

Since the origin $(0,0)$ is counted once, the total number of distinct intersection points is exactly three: $(0,0)$, $(-2,0)$, and $(0,-4)$. Thus, $p=0$ is a valid solution.

To obtain exactly three points, the circle could be tangent to one axis (providing one point) while intersecting the other axis at two distinct points (providing two points).

First, test tangency to the $x$-axis by setting $y=0$:

For tangency, the discriminant $D$ must be zero:

Now, verify $p=-1$ by checking the $y$-axis ($x=0$): The discriminant is $D=4^2-4(1)(1)=12$. Since $12>0$, there are two distinct intersection points on the $y$-axis. Therefore, $p=-1$ is a valid solution.

Next, test tangency to the $y$-axis by setting $x=0$:

For tangency, the discriminant must be zero:

Verify $p=-4$ by checking the $x$-axis ($y=0$): The discriminant is $D=2^2-4(1)(4)=-12$. Since the discriminant is negative, there are no real intersection points on the $x$-axis. This case fails to meet the requirement of three distinct points.

Through rigorous testing of the geometric constraints, we have determined that the values of $p$ that result in exactly three intersection points are $p=0$ and $p=-1$.

The final set of values is $\{0,-1\}$.