1. Write the circle in standard form

Given circle: $x^2+y^2+2x+4y-p=0$

Complete squares: $x^2+2x+y^2+4y=p$ $(x+1)^2-1+(y+2)^2-4=p$ $(x+1)^2+(y+2)^2=p+5$

So the circle has:

• Centre $(-1,-2)$
• Radius $r=\sqrt{p+5}$

For a real circle, we need: $p+5\ge 0$

2. Find intersections with the coordinate axes

The coordinate axes are:

• $x$-axis: $y=0$
• $y$-axis: $x=0$

On the $x$-axis

Put $y=0$ in the circle equation: $x^2+2x-p=0$ Discriminant: $\Delta_x=2^2-4(1)(-p)=4+4p=4(p+1)$

So:

• two distinct points if $p>-1$
• one point (tangent) if $p=-1$
• no real point if $p<-1$

On the $y$-axis

Put $x=0$ in the circle equation: $y^2+4y-p=0$ Discriminant: $\Delta_y=4^2-4(1)(-p)=16+4p=4(p+4)$

So:

• two distinct points if $p>-4$
• one point (tangent) if $p=-4$
• no real point if $p<-4$

3. Count total common points with the two axes

We need the circle and the coordinate axes to have exactly three common points in total.

Note that the two axes intersect at the origin $(0,0)$. If the circle passes through the origin, then a point on both axes is counted only once in the total set of common points.

Check whether origin lies on the circle: $0+0+0+0-p=0 \Rightarrow p=0$ So only for $p=0$, the origin is on the circle.

4. Casewise analysis

Case 1: Circle does not pass through origin ($p\ne 0$)

Then intersection points on $x$-axis and $y$-axis are distinct sets, so total number of common points is just the sum.

To get exactly 3 points, one axis must contribute 2 points and the other 1 point.

That happens when:

• $p=-1$:

  • On $x$-axis: one point
  • On $y$-axis: since $-1>-4$, two points
  • Total $=1+2=3$

• $p=-4$:

  • On $y$-axis: one point
  • On $x$-axis: since $-4<-1$, no point
  • Total $=1$, not 3

So from this case, only: $p=-1$

Case 2: Circle passes through origin ($p=0$)

Then:

• On $x$-axis: equation becomes $x^2+2x=0 \Rightarrow x(x+2)=0$ giving points $(0,0)$ and $(-2,0)$
• On $y$-axis: equation becomes $y^2+4y=0 \Rightarrow y(y+4)=0$ giving points $(0,0)$ and $(0,-4)$

Distinct common points are: $(0,0),\;(-2,0),\;(0,-4)$ So total number of distinct common points is 3.

Thus $p=0$ also works.

5. Conclusion

The required values of $p$ are: $p=-1,\;0$ Hence the number of values of $p$ is: $\boxed{2}$