Given pair of lines `2x^(2) +5xy +2y^(2) +4x +5y +a = 0` and the line `L: bx +y +5 = 0`. Then

Options

A. `a = 2`
B. `a =- 2`
C. There exists no circle which touches the pair of lines and the line L if `b = 5`.
D. There exists no circle which touches the pair of lines and the line L if `b =- 5`

Explanation

To solve the problem given the pair of lines represented by the equation \(2x^2 + 5xy + 2y^2 + 4x + 5y + a = 0\) and the line \(L: bx + y + 5 = 0\), we need to find the value of \(a\) and \(b\) under certain conditions. Here’s a step-by-step solution:

### Step 1: Determine the condition for a pair of lines
The equation \(2x^2 + 5xy + 2y^2 + 4x + 5y + a = 0\) represents a pair of lines if the determinant of the coefficients is equal to zero. The determinant \(D\) for the general conic section \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) is given by:

\[ D = \begin{vmatrix} A & \frac{B}{2} & \frac{D}{2} \\ \frac{B}{2} & C & \frac{E}{2} \\ \frac{D}{2} & \frac{E}{2} & F \end{vmatrix} \]

For our equation, we have:
- \(A = 2\)
- \(B = 5\)
- \(C = 2\)
- \(D = 4\)
- \(E = 5\)
- \(F = a\)

### Step 2: Set up the determinant
The determinant becomes:

\[ D = \begin{vmatrix} 2 & \frac{5}{2} & 2 \\ \frac{5}{2} & 2 & \frac{5}{2} \\ 2 & \frac{5}{2} & a \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant:

\[ D = 2 \begin{vmatrix} 2 & \frac{5}{2} \\ \frac{5}{2} & a \end{vmatrix} - \frac{5}{2} \begin{vmatrix} \frac{5}{2} & 2 \\ 2 & a \end{vmatrix} + 2 \begin{vmatrix} \frac{5}{2} & 2 \\ 2 & \frac{5}{2} \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & \frac{5}{2} \\ \frac{5}{2} & a \end{vmatrix} = 2a - \frac{25}{4} \)
2. \( \begin{vmatrix} \frac{5}{2} & 2 \\ 2 & a \end{vmatrix} = \frac{5a}{2} - 4 \)
3. \( \begin{vmatrix} \frac{5}{2} & 2 \\ 2 & \frac{5}{2} \end{vmatrix} = \frac{25}{4} - 4 = \frac{9}{4} \)

Substituting back into the determinant:

\[ D = 2(2a - \frac{25}{4}) - \frac{5}{2}(\frac{5a}{2} - 4) + 2(\frac{9}{4}) \]

### Step 4: Simplify the determinant
Expanding this gives:

\[ D = 4a - \frac{50}{4} - \frac{25a}{4} + 10 + \frac{18}{4} \]

Combining like terms:

\[ D = 4a - \frac{25a}{4} + 10 - \frac{32}{4} \]

This simplifies to:

\[ D = 4a - \frac{25a}{4} - \frac{22}{4} = 0 \]

### Step 5: Solve for \(a\)
To solve for \(a\), we multiply through by 4 to eliminate the fraction:

\[ 16a - 25a - 22 = 0 \implies -9a - 22 = 0 \implies 9a = 22 \implies a = \frac{22}{9} \]

### Step 6: Find \(b\)
Next, we check the line \(L: bx + y + 5 = 0\) to find \(b\) such that the lines are concurrent. The determinant must also equal zero for the lines to be concurrent:

\[ \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & 2 \\ b & 1 & 5 \end{vmatrix} = 0 \]

Calculating this determinant gives:

\[ 1(1 \cdot 5 - 2 \cdot 2) - 2(2 \cdot 5 - 2b) + 1(2 - 2b) = 0 \]

This simplifies to:

\[ 5 - 4 - 20 + 4b + 2 - 2b = 0 \implies 0 + 2b - 20 = 0 \implies 2b = 20 \implies b = 10 \]

### Final Answer:
Thus, the values of \(a\) and \(b\) are:

\[ a = \frac{22}{9}, \quad b = 10 \]