If a point P is moving such that the lengths
of the tangents drawn form P to the circles
`x^(2) + y^(2) + 8x + 12y + 15 = 0 `and
`x^(2) + y^(2) - 4 x - 6y - 12 = 0 ` are equal
then find the equation of the locus of P

Answer

To find the equation of the locus of point P from which the lengths of the tangents drawn to the two given circles are equal, we need to follow these steps:

### Step 1: Write the equations of the circles
The equations of the circles are given as:
1. \( C_1: x^2 + y^2 + 8x + 12y + 15 = 0 \)
2. \( C_2: x^2 + y^2 - 4x - 6y - 12 = 0 \)

### Step 2: Rearrange the equations to standard form
We can rearrange both equations into the standard form of a circle, which is \((x - h)^2 + (y - k)^2 = r^2\).

For \( C_1 \):
\[ x^2 + 8x + y^2 + 12y + 15 = 0 \]
Completing the square for \( x \) and \( y \):
\[ (x^2 + 8x + 16) + (y^2 + 12y + 36) = 1 \]
This simplifies to:
\[ (x + 4)^2 + (y + 6)^2 = 1 \]
Thus, the center is \((-4, -6)\) and the radius is \(1\).

For \( C_2 \):
\[ x^2 - 4x + y^2 - 6y - 12 = 0 \]
Completing the square for \( x \) and \( y \):
\[ (x^2 - 4x + 4) + (y^2 - 6y + 9) = 25 \]
This simplifies to:
\[ (x - 2)^2 + (y - 3)^2 = 25 \]
Thus, the center is \((2, 3)\) and the radius is \(5\).

### Step 3: Use the condition for equal tangents
The lengths of the tangents from point \( P(x, y) \) to the circles are equal if:
\[ \text{Length of tangent to } C_1 = \text{Length of tangent to } C_2 \]
This condition can be expressed using the radical axis:
\[ s_1 - s_2 = 0 \]
where \( s_1 \) and \( s_2 \) are the expressions for the circles.

### Step 4: Set up the equation \( s_1 - s_2 = 0 \)
Substituting the equations of the circles into \( s_1 - s_2 \):
\[ (x^2 + y^2 + 8x + 12y + 15) - (x^2 + y^2 - 4x - 6y - 12) = 0 \]
This simplifies to:
\[ 8x + 12y + 15 + 4x + 6y + 12 = 0 \]
Combining like terms:
\[ (8x + 4x) + (12y + 6y) + (15 + 12) = 0 \]
This gives:
\[ 12x + 18y + 27 = 0 \]

### Step 5: Final equation of the locus
Thus, the equation of the locus of point P is:
\[ 12x + 18y + 27 = 0 \]