One of the limit point of the coaxial system of circles containing `x^(2)+y^(2)-6x-6y+4=0, x^(2)+y^(2)-2x-4y+3=0`, is

Options

A. (-1, 1)
B. (-1, 2)
C. (-2, 1)
D. (-2, 2)

Answer

To solve the given problem, we need to find one of the limit points of the coaxial system of circles defined by the equations:

1. \( x^2 + y^2 - 6x - 6y + 4 = 0 \)
2. \( x^2 + y^2 - 2x - 4y + 3 = 0 \)

### Step 1: Rewrite the Circle Equations
First, we can rewrite the equations of the circles in standard form by completing the square.

**For the first circle:**
\[ x^2 - 6x + y^2 - 6y + 4 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x - 3)^2 - 9 + (y - 3)^2 - 9 + 4 = 0 \]
\[ (x - 3)^2 + (y - 3)^2 - 14 = 0 \]
\[ (x - 3)^2 + (y - 3)^2 = 14 \]
This circle has center \( (3, 3) \) and radius \( \sqrt{14} \).

**For the second circle:**
\[ x^2 - 2x + y^2 - 4y + 3 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 + 3 = 0 \]
\[ (x - 1)^2 + (y - 2)^2 - 2 = 0 \]
\[ (x - 1)^2 + (y - 2)^2 = 2 \]
This circle has center \( (1, 2) \) and radius \( \sqrt{2} \).

### Step 2: Set Up the Coaxial System
The equation of the coaxial system of circles can be expressed as:
\[ C_1 + \lambda C_2 = 0 \]
Where \(C_1\) and \(C_2\) are the equations of the circles.

Substituting the equations:
\[ (x^2 + y^2 - 6x - 6y + 4) + \lambda (x^2 + y^2 - 2x - 4y + 3) = 0 \]

### Step 3: Combine the Equations
Combining the equations:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - (6 + 2\lambda)x - (6 + 4\lambda)y + (4 + 3\lambda) = 0 \]

### Step 4: Find the Center and Radius
The center of the resulting circle can be calculated using:
\[ \text{Center} = \left( \frac{-(g)}{2}, \frac{-(f)}{2} \right) \]
Where \(g\) and \(f\) are the coefficients of \(x\) and \(y\) respectively.

From the combined equation:
- \(g = -(6 + 2\lambda)\)
- \(f = -(6 + 4\lambda)\)

Thus, the center is:
\[ \left( \frac{6 + 2\lambda}{2}, \frac{6 + 4\lambda}{2} \right) = (3 + \lambda, 3 + 2\lambda) \]

### Step 5: Calculate the Radius
The radius squared \(r^2\) is given by:
\[ r^2 = \left(3 + 2\lambda\right)^2 + \left(3 + \lambda\right)^2 - (4 + 3\lambda) \]

Expanding this:
\[ = (9 + 12\lambda + 4\lambda^2) + (9 + 6\lambda + \lambda^2) - (4 + 3\lambda) \]
\[ = 5\lambda^2 + 17\lambda + 14 \]

### Step 6: Set Radius to Zero for Limit Points
For limit points, set \(r^2 = 0\):
\[ 5\lambda^2 + 17\lambda + 14 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \(a = 5\), \(b = 17\), \(c = 14\):
\[ \lambda = \frac{-17 \pm \sqrt{17^2 - 4 \cdot 5 \cdot 14}}{2 \cdot 5} \]
\[ = \frac{-17 \pm \sqrt{289 - 280}}{10} \]
\[ = \frac{-17 \pm 3}{10} \]
Calculating the two possible values of \(\lambda\):
\[ \lambda_1 = \frac{-14}{10} = -1.4 \quad \text{and} \quad \lambda_2 = \frac{-20}{10} = -2 \]

### Step 8: Find the Limit Points
Substituting \(\lambda_1\) and \(\lambda_2\) back into the center coordinates:
1. For \(\lambda = -1.4\):
\[ \text{Center} = (3 - 1.4, 3 - 2.8) = (1.6, 0.2) \]
2. For \(\lambda = -2\):
\[ \text{Center} = (3 - 2, 3 - 4) = (1, -1) \]

Thus, one of the limit points of the coaxial system of circles is \( (1, -1) \).