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 defined by the equations \(x^2 + y^2 - 4 = 0\) and \(x^2 + y^2 - x - y = 0\), we will follow these steps:

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

### Step 2: Express the equations in standard form
The first circle \(C_1\) has a center at \((0, 0)\) and a radius of \(2\) (since \(r = \sqrt{4} = 2\)).

For the second circle \(C_2\):
\[ x^2 + y^2 - x - y = 0 \implies x^2 - x + y^2 - y = 0 \]
Completing the square for \(x\) and \(y\):
\[ \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{2} \]
This represents a circle with center \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and radius \(\frac{1}{\sqrt{2}}\).

### Step 3: Set up the equation for the coaxial system
The general equation for a coaxial system of circles is given by:
\[ C_1 + \lambda(C_2 - C_1) = 0 \]
Substituting the expressions for \(C_1\) and \(C_2\):
\[ (x^2 + y^2 - 4) + \lambda\left((x^2 + y^2 - x - y) - (x^2 + y^2 - 4)\right) = 0 \]
This simplifies to:
\[ (x^2 + y^2 - 4) + \lambda(-x - y + 4) = 0 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ x^2 + y^2 + \lambda(-x - y + 4) - 4 = 0 \]
This can be expressed as:
\[ x^2 + y^2 - \lambda x - \lambda y + 4\lambda - 4 = 0 \]

### Step 5: Identify coefficients for the radius condition
The general form of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Comparing coefficients, we have:
- \(2g = -\lambda\)
- \(2f = -\lambda\)
- \(c = 4\lambda - 4\)

### Step 6: Calculate the radius
The radius \(r\) of the circle is given by:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values:
\[ g = -\frac{\lambda}{2}, \quad f = -\frac{\lambda}{2}, \quad c = 4\lambda - 4 \]
Thus,
\[ r = \sqrt{\left(-\frac{\lambda}{2}\right)^2 + \left(-\frac{\lambda}{2}\right)^2 - (4\lambda - 4)} \]
This simplifies to:
\[ r = \sqrt{\frac{\lambda^2}{4} + \frac{\lambda^2}{4} - (4\lambda - 4)} = \sqrt{\frac{\lambda^2}{2} - 4\lambda + 4} \]

### Step 7: Set the radius to zero
Setting the radius to zero gives:
\[ \frac{\lambda^2}{2} - 4\lambda + 4 = 0 \]
Multiplying through by 2:
\[ \lambda^2 - 8\lambda + 8 = 0 \]

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

### Step 9: Find the limiting points
For \(\lambda = 4 + 2\sqrt{2}\) and \(\lambda = 4 - 2\sqrt{2}\), we can find the corresponding centers of the circles:
\[ \text{Center} = \left(-\frac{\lambda}{2}, -\frac{\lambda}{2}\right) \]
Calculating for both values of \(\lambda\):
1. For \(\lambda = 4 + 2\sqrt{2}\):
\[ \text{Center} = \left(-2 - \sqrt{2}, -2 - \sqrt{2}\right) \]
2. For \(\lambda = 4 - 2\sqrt{2}\):
\[ \text{Center} = \left(-2 + \sqrt{2}, -2 + \sqrt{2}\right) \]

### Conclusion
One of the limiting points of the coaxial system of circles is \((-2 - \sqrt{2}, -2 - \sqrt{2})\) or \((-2 + \sqrt{2}, -2 + \sqrt{2})\).