The value of 'p' so that `x^(2) + y^(2) + 8x + 10y + p = 0 ` is the equation of a circle of radius 7 units is ............ .

Answer

To find the value of 'p' such that the equation \(x^2 + y^2 + 8x + 10y + p = 0\) represents a circle with a radius of 7 units, we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ x^2 + y^2 + 8x + 10y + p = 0 \]
We can rearrange it to isolate the constant term:
\[ x^2 + y^2 + 8x + 10y = -p \]

### Step 2: Complete the square for the x terms
To complete the square for the \(x\) terms, we take \(x^2 + 8x\):
\[ x^2 + 8x = (x + 4)^2 - 16 \]
So we can rewrite the equation as:
\[ (x + 4)^2 - 16 + y^2 + 10y = -p \]

### Step 3: Complete the square for the y terms
Next, we complete the square for the \(y\) terms, \(y^2 + 10y\):
\[ y^2 + 10y = (y + 5)^2 - 25 \]
Now substituting this back into the equation gives:
\[ (x + 4)^2 - 16 + (y + 5)^2 - 25 = -p \]
This simplifies to:
\[ (x + 4)^2 + (y + 5)^2 - 41 = -p \]

### Step 4: Rearranging to standard circle form
Rearranging this, we have:
\[ (x + 4)^2 + (y + 5)^2 = 41 - p \]

### Step 5: Relate to the radius
For the equation to represent a circle, we know that the right side must equal the square of the radius. Given that the radius is 7, we have:
\[ 41 - p = 7^2 \]
Calculating \(7^2\):
\[ 7^2 = 49 \]
So we set up the equation:
\[ 41 - p = 49 \]

### Step 6: Solve for p
Now, we solve for \(p\):
\[ -p = 49 - 41 \]
\[ -p = 8 \]
Thus, multiplying both sides by -1 gives:
\[ p = -8 \]

### Final Answer
The value of \(p\) is \(-8\).

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