Value of p, of which are the equation `x^2+y^2-2px+4y-12=0` represent a circle of radius 5 units is

Options

A. 3
B. -3
C. BPTH (a)&(b)
D. neither (a) nor (b)

Answer

To find the value of \( p \) such that the equation \( x^2 + y^2 - 2px + 4y - 12 = 0 \) represents a circle of radius 5 units, we can follow these steps:

### Step 1: Identify the general form of the circle equation
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \( g \), \( f \), and \( c \) are constants.

### Step 2: Compare the given equation with the general form
The given equation is:
\[ x^2 + y^2 - 2px + 4y - 12 = 0 \]
From this, we can identify:
- \( 2g = -2p \) → \( g = -p \)
- \( 2f = 4 \) → \( f = 2 \)
- \( c = -12 \)

### Step 3: Use the radius formula
The radius \( r \) of the circle can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Given that the radius is 5 units, we can set up the equation:
\[ 5 = \sqrt{g^2 + f^2 - c} \]

### Step 4: Substitute the values of \( g \), \( f \), and \( c \)
Substituting \( g = -p \), \( f = 2 \), and \( c = -12 \) into the radius formula:
\[ 5 = \sqrt{(-p)^2 + 2^2 - (-12)} \]
This simplifies to:
\[ 5 = \sqrt{p^2 + 4 + 12} \]
\[ 5 = \sqrt{p^2 + 16} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 25 = p^2 + 16 \]

### Step 6: Solve for \( p^2 \)
Rearranging the equation:
\[ p^2 = 25 - 16 \]
\[ p^2 = 9 \]

### Step 7: Find the values of \( p \)
Taking the square root of both sides:
\[ p = \pm 3 \]

### Conclusion
The values of \( p \) that satisfy the equation are:
\[ p = 3 \quad \text{and} \quad p = -3 \]