If P and Q are the points of intersection of the circles `x^(2)+y^(2)+ 3x + 7y +2p-5=0` and `x^(2)+y^(2)+2x+2y+p^(2)=0`, then there is a circle passing through P and Q and (1, 1) for

Options

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

Answer

To solve the problem, we need to find the equation of the circle that passes through the points of intersection \( P \) and \( Q \) of the given circles, as well as the point \( (1, 1) \).

### Step 1: Write down the equations of the circles

The equations of the circles are:

1. \( x^2 + y^2 + 3x + 7y + 2p - 5 = 0 \) (Circle 1)
2. \( x^2 + y^2 + 2x + 2y + p^2 = 0 \) (Circle 2)

### Step 2: Subtract the equations to eliminate \( x^2 + y^2 \)

By subtracting the second equation from the first, we can eliminate \( x^2 + y^2 \):

\[ (3x + 7y + 2p - 5) - (2x + 2y + p^2) = 0 \]

This simplifies to:

\[ x + 5y + 2p - p^2 - 5 = 0 \]

### Step 3: Rearranging the equation

Rearranging gives us:

\[ x + 5y + 2p - p^2 = 5 \]

### Step 4: Express \( x \) in terms of \( y \) and \( p \)

From the rearranged equation, we can express \( x \):

\[ x = 5 - 5y - 2p + p^2 \]

### Step 5: Substitute \( x \) back into one of the original circle equations

We can substitute this expression for \( x \) back into one of the original circle equations (let's use Circle 1):

\[ (5 - 5y - 2p + p^2)^2 + y^2 + 3(5 - 5y - 2p + p^2) + 7y + 2p - 5 = 0 \]

### Step 6: Simplify the equation

This will yield a quadratic equation in \( y \). We can simplify this equation step by step to find the values of \( y \).

### Step 7: Find the points of intersection \( P \) and \( Q \)

Once we have the quadratic equation in \( y \), we can find the roots which will give us the \( y \)-coordinates of points \( P \) and \( Q \). We can then substitute these \( y \)-values back into the expression for \( x \) to find the corresponding \( x \)-coordinates.

### Step 8: Find the equation of the circle through points \( P \), \( Q \), and \( (1, 1) \)

Let the center of the circle be \( (h, k) \) and the radius be \( r \). The general equation of a circle is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

We will use the coordinates of points \( P \), \( Q \), and \( (1, 1) \) to set up a system of equations to solve for \( h \), \( k \), and \( r \).

### Step 9: Solve the system of equations

By substituting the coordinates of points \( P \), \( Q \), and \( (1, 1) \) into the circle equation, we will have three equations to solve for \( h \), \( k \), and \( r \).

### Step 10: Write the final equation of the circle

Once we have \( h \), \( k \), and \( r \), we can write the final equation of the circle.

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