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 find one of the limit points of the coaxial system of circles given by the equations \(x^2 + y^2 - 6x - 6y + 4 = 0\) and \(x^2 + y^2 - 2x - 4y + 3 = 0\), we will follow these steps:

### Step 1: Identify the equations of the circles
We denote the first circle as \(S_1\) and the second circle as \(S_2\):
- \(S_1: x^2 + y^2 - 6x - 6y + 4 = 0\)
- \(S_2: x^2 + y^2 - 2x - 4y + 3 = 0\)

### Step 2: Rewrite the equations in standard form
To find the centers and radii of the circles, we can complete the square for each equation.

**For \(S_1\):**
\[ x^2 - 6x + y^2 - 6y + 4 = 0 \]
Completing the square:
\[ (x - 3)^2 - 9 + (y - 3)^2 - 9 + 4 = 0 \implies (x - 3)^2 + (y - 3)^2 = 14 \]
Thus, the center of \(S_1\) is \((3, 3)\) and the radius is \(\sqrt{14}\).

**For \(S_2\):**
\[ x^2 - 2x + y^2 - 4y + 3 = 0 \]
Completing the square:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 + 3 = 0 \implies (x - 1)^2 + (y - 2)^2 = 2 \]
Thus, the center of \(S_2\) is \((1, 2)\) and the radius is \(\sqrt{2}\).

### Step 3: Set up the coaxial system equation
The coaxial system of circles can be expressed as:
\[ S_1 + \lambda(S_1 - S_2) = 0 \]
This gives us:
\[ S_1 - S_2 = 0 \implies (x^2 + y^2 - 6x - 6y + 4) - (x^2 + y^2 - 2x - 4y + 3) = 0 \]
Simplifying this:
\[ -6x + 2x - 6y + 4y + 4 - 3 = 0 \implies -4x - 2y + 1 = 0 \]
Thus, we have:
\[ 4x + 2y = 1 \implies 2x + y = \frac{1}{2} \]

### Step 4: Find the limiting points
The limiting points can be found by substituting values of \(\lambda\) into the center coordinates derived from the coaxial system. The centers can be expressed as:
\[ C(\lambda) = (3 + 2\lambda, 3 + \lambda) \]

Setting the radius to zero for the limiting points:
\[ g^2 + f^2 - c = 0 \]
Substituting \(g = 3 + 2\lambda\), \(f = 3 + \lambda\), and \(c = 4 + \lambda\):
\[ (3 + 2\lambda)^2 + (3 + \lambda)^2 - (4 + \lambda) = 0 \]
Expanding and simplifying:
\[ 9 + 12\lambda + 4\lambda^2 + 9 + 6\lambda + 4 + 2\lambda - 4 - \lambda = 0 \]
Combining like terms:
\[ 5\lambda^2 + 17\lambda + 14 = 0 \]
Factoring or using the quadratic formula gives:
\[ \lambda = -2 \quad \text{or} \quad \lambda = -\frac{7}{5} \]

### Step 5: Calculate the limiting points
1. For \(\lambda = -2\):
\[ C(-2) = (3 + 2(-2), 3 + (-2)) = (3 - 4, 3 - 2) = (-1, 1) \]

2. For \(\lambda = -\frac{7}{5}\):
\[ C\left(-\frac{7}{5}\right) = \left(3 + 2\left(-\frac{7}{5}\right), 3 - \frac{7}{5}\right) = \left(3 - \frac{14}{5}, 3 - \frac{7}{5}\right) = \left(\frac{15 - 14}{5}, \frac{15 - 7}{5}\right) = \left(\frac{1}{5}, \frac{8}{5}\right) \]

### Conclusion
The limiting points of the coaxial system of circles are \((-1, 1)\) and \(\left(\frac{1}{5}, \frac{8}{5}\right)\). Thus, one of the limit points is \((-1, 1)\).