Let the equation x² + y² + px + (1 - p)y + 5 = 0 represent circles of varying radius r ∈ (0, 5]. The

Q: Wait, why do we have to split the problem into two cases for $p = \pm \sqrt{q}$?

Great question! The inequality we derived is $9.5 < p^2 - p \le 59.5$. Notice the term $-p$. If $p$ is positive, we subtract $\sqrt{q}$. If $p$ is negative, we add $\sqrt{q}$. Because the value of the expression $p^2 - p$ changes significantly based on the sign of $p$, we cannot treat it as a single case. We must analyze both scenarios to capture all possible integer values of $q$.

Q: Can we just square the inequality $9.5 < p^2 - p \le 59.5$ directly to solve for $p$?

Be careful! Squaring an inequality is only safe when all parts are positive. While $p^2 - p$ is indeed greater than $9.5$ (which is positive), the expression $p^2 - p$ itself can be negative for certain values of $p$ (specifically between $0$ and $1$). However, since our inequality forces $p^2 - p > 9.5$, we are safely in the positive region. But the most robust way to solve this is to substitute $p = \pm \sqrt{q}$ and solve for $q$ directly, as the question asks for the number of integer values of $q$.

Q: How do I know if I should include the endpoints when counting the integers?

Always look back at the original constraint: $r \in (0, 5]$. The parenthesis $(0$ means $r$ cannot be $0$, but the bracket $5]$ means $r$ can be exactly $5$. Therefore, when you solve the inequality $r \le 5$, you must include the value that makes $r=5$ (which corresponds to $q=67$ in our final set). Always check your boundary conditions!

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 $S = \{q : q = p^2 	ext{ and } q 	ext{ is an integer } \}$ is .

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of a Family of Circles

The equation $x^2 + y^2 + px + (1 - p)y + 5 = 0$ represents a family of circles dependent on the parameter $p$. Our objective is to determine the number of integer values of $q = p^2$ such that the radius $r$ satisfies the condition $0 < r \le 5$.

### Decoding the DNA of the Circle

To find the radius, we compare the given equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$. By matching coefficients, we identify:
$g = \frac{p}{2}$, $f = \frac{1 - p}{2}$, and $c = 5$.

The radius $r$ is defined by the formula $r = \sqrt{g^2 + f^2 - c}$. Substituting our values, we obtain:

$$r = \sqrt{\left(\frac{p}{2}\right)^2 + \left(\frac{1 - p}{2}\right)^2 - 5}$$

Expanding the squares and simplifying under a common denominator, we arrive at the expression for the radius:

$$r = \frac{\sqrt{2p^2 - 2p - 19}}{2}$$

### The Constraint and the Algebraic Dance

We are given the constraint $0 < r \le 5$. Substituting our expression for $r$ and multiplying by $2$, we get:

$$0 < \sqrt{2p^2 - 2p - 19} \le 10$$

Squaring all parts of the inequality to eliminate the square root yields:

$$0 < 2p^2 - 2p - 19 \le 100$$

Adding $19$ to all parts and dividing by $2$, we reach the governing inequality for $p$:

$$9.5 < p^2 - p \le 59.5$$

### The Bifurcation of $p$

Since we seek integer values of $q = p^2$, we must consider $p = \pm \sqrt{q}$. We analyze the two resulting cases separately.

**Case 1: $p = \sqrt{q}$**
Substituting into the inequality, we get $9.5 < q - \sqrt{q} \le 59.5$. Testing integer values for $q$:
For $q = 13$, $13 - \sqrt{13} \approx 9.39$ (too small).
For $q = 14$, $14 - \sqrt{14} \approx 10.26$ (valid).
For the upper bound, $q = 67$ gives $67 - \sqrt{67} \approx 58.81$ (valid), while $q = 68$ exceeds $59.5$.
Thus, $q \in \{14, 15, \dots, 67\}$.

**Case 2: $p = -\sqrt{q}$**
Substituting into the inequality, we get $9.5 < q + \sqrt{q} \le 59.5$. Testing integer values for $q$:
For $q = 6$, $6 + \sqrt{6} \approx 8.45$ (too small).
For $q = 7$, $7 + \sqrt{7} \approx 9.64$ (valid).
For the upper bound, $q = 52$ gives $52 + \sqrt{52} \approx 59.21$ (valid), while $q = 53$ exceeds $59.5$.
Thus, $q \in \{7, 8, \dots, 52\}$.

### Final Calculation

The union of the two sets of valid integers for $q$ is $\{7, 8, \dots, 67\}$.

To find the total count of these integers, we use the formula for the number of terms:
$67 - 7 + 1 = 61$.

There are exactly **61** integer values of $q$ that satisfy the given conditions.

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

Enter Numerical Value:

Visualized Solution

Analyze the Circle Equation

Identify $g, f, and c$

Apply the Radius Formula

Simplify the Radius Expression

Set up the Inequality

Square and Isolate $p$ terms

Introduce $q = p^{2}$

Solve Case 1: $p = q$

Solve Case 2: $p = - q$

Combine Sets and Count

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Geometry of a Family of Circles

Decoding the DNA of the Circle

The Constraint and the Algebraic Dance

The Bifurcation of $p$

Final Calculation

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Enter Numerical Value:

Visualized Solution

Analyze the Circle Equation

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g,f, and c

Apply the Radius Formula

Simplify the Radius Expression

Set up the Inequality

Square and Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p terms

Introduce .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q=p2

Solve Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=q​

Solve Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=−q​