Q.2 Find the complete set of values of $p$ for which the equation describes a circle in the $xy$-plane.

The given equation is:

$$x^2 - 2px + y^2 - 6y - p^2 + 8p + 9 = 0$$

Answer Choices

• $p < -\frac{9}{4}$
• $0 < p < 4$
• $-1 < p < 9$
• $p < 0 \text{ or } p > 4$
• $p < -1 \text{ or } p > 9$
• all real values of $p$

Solution

We are given the equation:

$$x^2 - 2px + y^2 - 6y - p^2 + 8p + 9 = 0$$

To describe a circle, the equation must be in the form:

$$(x - h)^2 + (y - k)^2 = r^2$$

where $(h, k)$ is the center and $r$ is the radius, and the radius $r$ must be positive. Let's try to complete the square for both the $x$-terms and the $y$-terms.

Step 1: Complete the square for $x$-terms

The $x$-terms are $x^2 - 2px$. Completing the square gives:

$$x^2 - 2px = (x - p)^2 - p^2$$

Step 2: Complete the square for $y$-terms

The $y$-terms are $y^2 - 6y$. Completing the square gives:

$$y^2 - 6y = (y - 3)^2 - 9$$

Step 3: Substitute into the equation

Substituting these results into the original equation:

$$(x - p)^2 - p^2 + (y - 3)^2 - 9 - p^2 + 8p + 9 = 0$$

Simplifying this expression:

$$(x - p)^2 + (y - 3)^2 = 2p^2 - 8p$$

Step 4: Set the condition for the radius

For the equation to describe a circle, the expression on the right-hand side must be positive, since it represents $r^2$, the square of the radius. Thus, we need:

$$2p^2 - 8p > 0$$

Factoring the quadratic gives:

$$2p(p - 4) > 0$$

This inequality is satisfied when $p > 4$ or $p < 0$.

Step 5: Conclusion

The equation describes a circle for $p > 4$ or $p < 0$. Therefore, the correct answer is: