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Pensiero

Solving circle and axes intersection problem

I need to figure out for how many values of p the circle intersects the x and y axes at exactly three points. The axes are the lines x=0 and y=0. This requires considering cases where the circle is tangent to one axis or passes through the origin. The circle could intersect the axes at up to four points, so three common points typically means one tangency and two intersection points, or passing through the origin.

Interpreting the common points issue

I’m thinking about how "common points" are defined: The circle intersects both axes, but the origin counts as the same point for both axes. To have exactly 3 distinct points, the circle must either be tangent to one axis and intersect the other in two, or pass through the origin, leaving only 3 distinct points. I'm considering both cases carefully, especially when tangency happens at the origin.

Clarifying circle tangency cases

I’m thinking about tangency: If the circle is tangent to one axis, then the distance from the center to that axis equals the radius. For the origin to be a common point, the circle would need to pass through the origin. Otherwise, in the case of tangency, we'd have 3 distinct points: one from the tangent axis and two from the other axis.

Circle passing through origin

If the circle intersects both axes at two points and passes through the origin, then the union of those points would have a total of four points. However, since the origin is counted once for both axes, it reduces the count to three distinct points. The origin counts as a common intersection point for both axes.

Analyzing circle p values for 3 points

For a circle to intersect both axes in three distinct points, there are two key scenarios:

Tangent to one axis, intersecting the other in two points.
Passing through the origin, intersecting both axes.

After testing with specific values, we find that p = 0 and p = -1 both yield three points. Further analysis shows there are no other valid p values resulting in exactly three points.

Summarizing p values for 3 points

After reviewing different cases, I found that the circle intersects the axes in three points only when p = -1 (tangent to the x-axis) and p = 0 (passes through the origin). Other values lead to different intersection counts, and degenerate cases do not give three intersections. Thus, there are exactly two such p values.

Thought for 2m 43s

Risposta

Answer: 2 values — $p=-1$ and $p=0$ .

Here are three clean ways to see it.

1) Discriminant counting (pure algebra)

Intersections with the axes come from setting $y=0$ or $x=0$ .

On the $x$ -axis: $x^2+2x-p=0$ .
Discriminant $\Delta_x=4(1+p)$ :
• two points if $p>-1$
• one (tangent) if $p=-1$
• none if $p<-1$
On the $y$ -axis: $y^2+4y-p=0$ .
Discriminant $\Delta_y=4(4+p)$ :
• two points if $p>-4$
• one (tangent) if $p=-4$
• none if $p<-4$

Total distinct common points $=\#x\text{-points}+\#y\text{-points}-\delta$ , where $\delta=1$ only if the origin $(0,0)$ lies on the circle (because it would be counted on both axes).
$(0,0)$ lies on the circle $\iff p=0$ .

Check the ranges:

$p=-1$ : $1$ point on $x$ -axis (tangent) + $2$ points on $y$ -axis $=3$ .
$p=0$ : $2$ points on each axis, but the origin is common to both, so $2+2-1=3$ .
All other $p$ give $0,1,2,$ or $4$ points.

So exactly three points occur for $p=-1$ and $p=0$ .

2) Geometry (center–radius picture)

Complete the square:

x^2+y^2+2x+4y-p=0 \;\Longleftrightarrow\; (x+1)^2+(y+2)^2=r^2,\quad r^2=p+5.

Center $C(-1,-2)$ . Distances from $C$ to the axes:

d(\text{x-axis})=|{-2}|=2,\qquad d(\text{y-axis})=|{-1}|=1.

To get three axis-intersections, either:

Tangent to one axis and secant to the other.
Only possible as tangent to the x-axis (need $r=2\Rightarrow p=-1$ ); then $r>1$ so it cuts the $y$ -axis in two points → total $3$ .
Pass through the origin (which sits on both axes), plus one more point on each axis.
Distance $CO=\sqrt{(-1)^2+(-2)^2}=\sqrt5\Rightarrow r^2=5\Rightarrow p=0$ .
Then you get $(-2,0),(0,0),(0,-4)$ : three distinct points.

No other geometric configuration gives exactly three.

3) Quick case table

\begin{array}{c|c|c|c} p & \#(x\text{-axis}) & \#(y\text{-axis}) & \text{Total distinct} \\ \hline p<-4 & 0 & 0 & 0\\ p=-4 & 0 & 1 & 1\\ -4<p<-1 & 0 & 2 & 2\\ p=-1 & 1 & 2 & 3\\ -1<p<0 & 2 & 2 & 4\\ p=0 & 2 & 2 & 3\ (\text{origin counted once})\\ p>0 & 2 & 2 & 4 \end{array}

Hence, exactly three occurs only for $p=-1$ and $p=0$ → 2 values.

I want you to solve this problem in as many ways a...