One of the limiting points of the co-axial system of circles containing the circles `x^(2)+y^(2)-4=0andx^(2)+y^(2)-x-y=0` is

Options

A. `(sqrt2,sqrt2)`
B. `(-sqrt2,sqrt2)`
C. `(-sqrt2-sqrt2)`
D. None of these

Answer

To find one of the limiting points of the coaxial system of circles containing the circles \( x^2 + y^2 - 4 = 0 \) and \( x^2 + y^2 - x - y = 0 \), we will follow these steps:

### Step 1: Identify the equations of the circles
The given circles are:
1. Circle 1: \( C_1: x^2 + y^2 - 4 = 0 \)
2. Circle 2: \( C_2: x^2 + y^2 - x - y = 0 \)

### Step 2: Write the general equation for the coaxial system
The general equation for the family of coaxial circles containing two circles is given by:
\[ C_1 + \lambda (C_1 - C_2) = 0 \]
Substituting the circles into this equation:
\[ (x^2 + y^2 - 4) + \lambda \left( (x^2 + y^2 - 4) - (x^2 + y^2 - x - y) \right) = 0 \]
This simplifies to:
\[ x^2 + y^2 - 4 + \lambda (x + y - 4) = 0 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ x^2 + y^2 + \lambda (x + y) - 4(1 + \lambda) = 0 \]

### Step 4: Identify coefficients for radius calculation
From the general form of the circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we can identify:
- \( g = \frac{\lambda}{2} \)
- \( f = \frac{\lambda}{2} \)
- \( c = -4(1 + \lambda) \)

### Step 5: Calculate the radius
The radius \( r \) of the circle is given by:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values of \( g \), \( f \), and \( c \):
\[ r = \sqrt{\left(\frac{\lambda}{2}\right)^2 + \left(\frac{\lambda}{2}\right)^2 + 4(1 + \lambda)} \]
This simplifies to:
\[ r = \sqrt{\frac{\lambda^2}{4} + \frac{\lambda^2}{4} + 4 + 4\lambda} = \sqrt{\frac{\lambda^2}{2} + 4 + 4\lambda} \]

### Step 6: Set the radius to zero to find limiting points
To find the limiting points, we set the radius \( r \) to zero:
\[ \frac{\lambda^2}{2} + 4 + 4\lambda = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ \lambda^2 + 8\lambda + 8 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \lambda = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{-8 \pm \sqrt{64 - 32}}{2} = \frac{-8 \pm \sqrt{32}}{2} \]
This simplifies to:
\[ \lambda = \frac{-8 \pm 4\sqrt{2}}{2} = -4 \pm 2\sqrt{2} \]

### Step 8: Find the limiting points
The limiting points are given by the center of the circle when \( r = 0 \):
\[ \text{Center} = \left(-\frac{g}{2}, -\frac{f}{2}\right) = \left(-\frac{\lambda}{4}, -\frac{\lambda}{4}\right) \]
Substituting the values of \( \lambda \):
1. For \( \lambda = -4 + 2\sqrt{2} \):
\[ \text{Center} = \left( -\frac{-4 + 2\sqrt{2}}{4}, -\frac{-4 + 2\sqrt{2}}{4} \right) = \left( 1 - \frac{\sqrt{2}}{2}, 1 - \frac{\sqrt{2}}{2} \right) \]
2. For \( \lambda = -4 - 2\sqrt{2} \):
\[ \text{Center} = \left( -\frac{-4 - 2\sqrt{2}}{4}, -\frac{-4 - 2\sqrt{2}}{4} \right) = \left( 1 + \frac{\sqrt{2}}{2}, 1 + \frac{\sqrt{2}}{2} \right) \]

### Conclusion
Thus, one of the limiting points of the coaxial system of circles is:
\[ \left( 1 - \frac{\sqrt{2}}{2}, 1 - \frac{\sqrt{2}}{2} \right) \]
or
\[ \left( 1 + \frac{\sqrt{2}}{2}, 1 + \frac{\sqrt{2}}{2} \right) \]