Question

Let the equation $x^2 + y^2 + px + (1 - p)y + 5 = 0$ represent circles of varying radius $r \in (0, 5]$. Then the number of elements in the set $S = \{q : q = p^2 \text{ and } q \text{ is an integer}\}$ is _________.

Hint:

Remember that for a real circle to exist, $g^2 + f^2 - c$ must be strictly positive. The number of integers in a range $[a, b]$ is $\lfloor b \rfloor - \lceil a \rceil + 1$.

Answer 1

Correct Answer: 61

Step 1: Understanding the Concept:
The radius $r$ of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $r = \sqrt{g^2 + f^2 - c}$. We need to apply the constraint $0<r^2 \leq 25$ to find the range for $p^2$.
Step 2: Detailed Explanation:
For the given circle $x^2 + y^2 + px + (1 - p)y + 5 = 0$:
$g = p/2, f = (1-p)/2, c = 5$.
$r^2 = g^2 + f^2 - c = \frac{p^2}{4} + \frac{(1-p)^2}{4} - 5 = \frac{p^2 + 1 - 2p + p^2 - 20}{4} = \frac{2p^2 - 2p - 19}{4}$.
Given $0<r \leq 5 \implies 0<r^2 \leq 25$.
1. $\frac{2p^2 - 2p - 19}{4}>0 \implies 2p^2 - 2p - 19>0$.
Roots are $\frac{2 \pm \sqrt{4 + 152}}{4} = \frac{1 \pm \sqrt{39}}{2} \approx 3.62, -2.62$.
So $p \in (-\infty, -2.62) \cup (3.62, \infty)$.
2. $\frac{2p^2 - 2p - 19}{4} \leq 25 \implies 2p^2 - 2p - 19 \leq 100 \implies 2p^2 - 2p - 119 \leq 0$.
Roots are $\frac{2 \pm \sqrt{4 + 952}}{4} = \frac{1 \pm \sqrt{239}}{2} \approx 8.23, -7.23$.
So $p \in [-7.23, 8.23]$.
Combined range for $p$: $[-7.23, -2.62) \cup (3.62, 8.23]$.
Now, find the range of $q = p^2$:
From $p \in [-7.23, -2.62)$, $p^2 \in (6.86, 52.27]$.
From $p \in (3.62, 8.23]$, $p^2 \in (13.10, 67.73]$.
Union of $p^2$ values is $(6.86, 67.73]$.
Integers in this range are $\{7, 8, 9, ..., 67\}$.
Total count $= 67 - 7 + 1 = 61$.
Step 3: Final Answer:
The number of elements in the set $S$ is 61.