If P and Q are the points of intersection of the circles `x^2+y^2+3x+7y+2p=0` and `x^2+y^2+2x+2y-p^2=0` then there is a circle passing through P,Q and (1,1) for

Options

A. all values of p
B. all except one value of p
C. all except two values of p
D. exactly one value of p

Explanation

To solve the problem, we need to determine the values of \( p \) for which there exists a circle passing through the points of intersection \( P \) and \( Q \) of the two given circles, as well as the point \( (1, 1) \).

### Step-by-Step Solution:

1. **Write the equations of the circles**:
The equations of the circles are given as:
\[ C_1: x^2 + y^2 + 3x + 7y + 2p = 0 \]
\[ C_2: x^2 + y^2 + 2x + 2y - p^2 = 0 \]

2. **Form the general equation of the family of circles**:
The general equation of a family of circles can be expressed as:
\[ s_1 + \lambda s_2 = 0 \]
where \( s_1 \) and \( s_2 \) are the equations of the circles. Thus, we have:
\[ (x^2 + y^2 + 3x + 7y + 2p) + \lambda (x^2 + y^2 + 2x + 2y - p^2) = 0 \]

3. **Substitute the point (1, 1)**:
Since the circle passes through the point \( (1, 1) \), we substitute \( x = 1 \) and \( y = 1 \) into the equation:
\[ 1^2 + 1^2 + 3(1) + 7(1) + 2p + \lambda(1^2 + 1^2 + 2(1) + 2(1) - p^2) = 0 \]
Simplifying this, we get:
\[ 1 + 1 + 3 + 7 + 2p + \lambda(2 + 2 + 2 - p^2) = 0 \]
\[ 12 + 2p + \lambda(6 - p^2) = 0 \]

4. **Rearranging the equation**:
Rearranging gives us:
\[ 2p + 6\lambda - \lambda p^2 + 12 = 0 \]

5. **Set \( \lambda = -1 \)**:
To find conditions on \( p \), we can set \( \lambda = -1 \):
\[ 2p + 6(-1) - (-1)p^2 + 12 = 0 \]
Simplifying this:
\[ 2p - 6 + p^2 + 12 = 0 \]
\[ p^2 + 2p + 6 = 0 \]

6. **Finding the discriminant**:
The discriminant \( D \) of the quadratic equation \( p^2 + 2p + 6 = 0 \) is given by:
\[ D = b^2 - 4ac = 2^2 - 4(1)(6) = 4 - 24 = -20 \]
Since the discriminant is negative, there are no real solutions for \( p \).

7. **Conclusion**:
Since there are no restrictions on \( p \) from the quadratic equation, we conclude that the circle can pass through points \( P \), \( Q \), and \( (1, 1) \) for all values of \( p \) except when \( \lambda = -1 \) because this would not represent a valid circle.

### Final Answer:
Thus, the answer is that there is a circle passing through \( P \), \( Q \), and \( (1, 1) \) for all values of \( p \) except \( p = -1 \).