Key Concepts and Formulas

• The equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$, where $(-g, -f)$ is the center and $\sqrt{g^2 + f^2 - c}$ is the radius.
• The equation of the family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$, where $\lambda$ is a parameter.
• If a point $(x_1, y_1)$ lies on a circle $S = 0$, then $S(x_1, y_1) = 0$.

Step-by-Step Solution

Step 1: Define the given circles

Let the equations of the two circles be $S_1 \equiv x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ $S_2 \equiv x^2 + y^2 + 2x + 2y - p^2 = 0$

Step 2: Form the equation of the family of circles

The equation of the family of circles passing through the intersection of $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$ Substituting the expressions for $S_1$ and $S_2$, we get $(x^2 + y^2 + 3x + 7y + 2p - 5) + \lambda (x^2 + y^2 + 2x + 2y - p^2) = 0$ $(1 + \lambda)x^2 + (1 + \lambda)y^2 + (3 + 2\lambda)x + (7 + 2\lambda)y + (2p - 5 - \lambda p^2) = 0$

Step 3: Use the condition that the circle passes through (1, 1)

Since the circle passes through the point $(1, 1)$, we substitute $x = 1$ and $y = 1$ into the equation of the family of circles: $(1 + \lambda)(1)^2 + (1 + \lambda)(1)^2 + (3 + 2\lambda)(1) + (7 + 2\lambda)(1) + (2p - 5 - \lambda p^2) = 0$ $1 + \lambda + 1 + \lambda + 3 + 2\lambda + 7 + 2\lambda + 2p - 5 - \lambda p^2 = 0$ $7 + 6\lambda + 2p - \lambda p^2 = 0$ $\lambda(6 - p^2) = -7 - 2p$

Step 4: Solve for $\lambda$

If $6 - p^2 \neq 0$, we can solve for $\lambda$: $\lambda = \frac{-7 - 2p}{6 - p^2}$

Step 5: Analyze the condition for existence of $\lambda$

The value of $\lambda$ exists as long as $6 - p^2 \neq 0$, which means $p^2 \neq 6$, so $p \neq \pm \sqrt{6}$. Thus, there are two values of $p$ for which $\lambda$ does not exist. Therefore, for all values of $p$ except two, there is a circle passing through $P, Q$ and $(1, 1)$.

Common Mistakes & Tips

• Remember to consider the case when the coefficient of $\lambda$ is zero.
• Be careful with algebraic manipulations to avoid errors.
• Understanding the concept of the family of circles is crucial for solving this type of problem.

Summary

We used the concept of the family of circles passing through the intersection of two circles. We then applied the condition that the circle passes through the point (1, 1) to find a relationship between $\lambda$ and $p$. By analyzing the condition for the existence of $\lambda$, we determined that there are two values of $p$ for which the circle does not exist. Therefore, there is a circle passing through $P, Q$ and $(1, 1)$ for all except two values of $p$.

Final Answer

The final answer is \boxed{all except two values of $p$}, which corresponds to option (B).