The limiting points of the system of circles represented by the equation `2(x^(2)+y^(2))+lambda x+(9)/(2)=0`, are

Options

A. `(pm(3)/(2),0)`
B. `(0,0)and((9)/(2),0)`
C. `(pm(9)/(2),0)`
D. `(pm3,0)`

Answer

To find the limiting points of the system of circles represented by the equation

\[ 2(x^2 + y^2) + \lambda x + \frac{9}{2} = 0, \]

we will follow these steps:

### Step 1: Rewrite the equation in standard form
First, we divide the entire equation by 2 to simplify:

\[ x^2 + y^2 + \frac{\lambda}{2} x + \frac{9}{4} = 0. \]

### Step 2: Identify coefficients
In the standard circle equation \(x^2 + y^2 + 2gx + 2fy + c = 0\), we can identify:

- \(2g = \frac{\lambda}{2}\) → \(g = \frac{\lambda}{4}\)
- \(2f = 0\) → \(f = 0\)
- \(c = \frac{9}{4}\)

### Step 3: Determine the center and radius
The center of the circle is given by \((-g, -f)\) and the radius \(r\) is given by:

\[ r = \sqrt{g^2 + f^2 - c}. \]

Substituting the values we found:

- Center: \(\left(-\frac{\lambda}{4}, 0\right)\)
- Radius:

\[ r = \sqrt{\left(\frac{\lambda}{4}\right)^2 + 0^2 - \frac{9}{4}} = \sqrt{\frac{\lambda^2}{16} - \frac{9}{4}}. \]

### Step 4: Set the radius to zero for limiting points
To find the limiting points, we set the radius \(r\) to zero:

\[ \frac{\lambda^2}{16} - \frac{9}{4} = 0. \]

### Step 5: Solve for \(\lambda\)
Multiplying through by 16 to eliminate the fraction:

\[ \lambda^2 - 36 = 0. \]

This factors to:

\[ (\lambda - 6)(\lambda + 6) = 0. \]

Thus, we find:

\[ \lambda = 6 \quad \text{or} \quad \lambda = -6. \]

### Step 6: Find the limiting points
Substituting \(\lambda\) back into the center coordinates:

For \(\lambda = 6\):

\[ \text{Center} = \left(-\frac{6}{4}, 0\right) = \left(-\frac{3}{2}, 0\right). \]

For \(\lambda = -6\):

\[ \text{Center} = \left(-\frac{-6}{4}, 0\right) = \left(\frac{3}{2}, 0\right). \]

### Final Result
The limiting points of the system of circles are:

\[ \left(-\frac{3}{2}, 0\right) \quad \text{and} \quad \left(\frac{3}{2}, 0\right). \]

Thus, the limiting points can be expressed as:

\[ \left(\pm \frac{3}{2}, 0\right). \]