The point of contact of the circles `x^(2)+y^(2)-4x+6y-3=0 and x^(2)+y^(2)+16x+6y+37=0` is

Options

A. `(-8,-3)`
B. (2,-3)
C. (-2,-3)
D. none

Answer

To find the point of contact of the two circles given by the equations:

1. \( x^2 + y^2 - 4x + 6y - 3 = 0 \) (let's call this Circle 1)
2. \( x^2 + y^2 + 16x + 6y + 37 = 0 \) (let's call this Circle 2)

we will follow these steps:

### Step 1: Write the equations of the circles in standard form
To do this, we will complete the square for both circles.

**Circle 1:**
\[ x^2 - 4x + y^2 + 6y - 3 = 0 \]
Completing the square for \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
Completing the square for \(y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]
Substituting back into the equation:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 3 = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y + 3)^2 = 16 \]
So, Circle 1 has center \( (2, -3) \) and radius \( 4 \).

**Circle 2:**
\[ x^2 + 16x + y^2 + 6y + 37 = 0 \]
Completing the square for \(x\):
\[ x^2 + 16x = (x + 8)^2 - 64 \]
Completing the square for \(y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]
Substituting back into the equation:
\[ (x + 8)^2 - 64 + (y + 3)^2 - 9 + 37 = 0 \]
This simplifies to:
\[ (x + 8)^2 + (y + 3)^2 = 36 \]
So, Circle 2 has center \( (-8, -3) \) and radius \( 6 \).

### Step 2: Find the equation of the common tangent
The equation of the common tangent to two circles can be found using the formula:
\[ S_1 - S_2 = 0 \]
where \( S_1 \) and \( S_2 \) are the equations of the circles.

Substituting the equations:
\[ (x^2 + y^2 - 4x + 6y - 3) - (x^2 + y^2 + 16x + 6y + 37) = 0 \]
This simplifies to:
\[ -4x + 6y - 3 - 16x - 6y - 37 = 0 \]
Combining like terms:
\[ -20x - 40 = 0 \]
Thus:
\[ 20x = -40 \implies x = -2 \]

### Step 3: Find the y-coordinate of the point of contact
Since we have \( x = -2 \), we can substitute this value back into either circle's equation to find the corresponding \( y \)-coordinate. Using Circle 1's equation:
\[ (-2)^2 + y^2 - 4(-2) + 6y - 3 = 0 \]
This simplifies to:
\[ 4 + y^2 + 8 + 6y - 3 = 0 \implies y^2 + 6y + 9 = 0 \]
Factoring gives:
\[ (y + 3)^2 = 0 \implies y = -3 \]

### Final Answer
Thus, the point of contact of the two circles is:
\[ \boxed{(-2, -3)} \]