If `3x + b_(1)y + 5 =0` and `4x + b_(2)y + 10 = 0` cut the x-axis and y-axis in four concyclic, then the value of `b_(1)b_(2)` is

Options

A. (a)15
B. (b)30
C. (c)20
D. (d)12

Answer

To solve the problem, we need to find the value of \( b_1 b_2 \) given the two equations of lines that intersect the x-axis and y-axis at four concyclic points.

### Step-by-Step Solution:

1. **Identify the Equations**:
We have two equations:
\[ 3x + b_1y + 5 = 0 \quad \text{(Equation 1)} \]
\[ 4x + b_2y + 10 = 0 \quad \text{(Equation 2)} \]

2. **Find the Intercepts**:
- For Equation 1, to find the x-intercept, set \( y = 0 \):
\[ 3x + 5 = 0 \implies x = -\frac{5}{3} \]
The x-intercept is \( (-\frac{5}{3}, 0) \).
- To find the y-intercept, set \( x = 0 \):
\[ b_1y + 5 = 0 \implies y = -\frac{5}{b_1} \]
The y-intercept is \( (0, -\frac{5}{b_1}) \).

- For Equation 2, to find the x-intercept, set \( y = 0 \):
\[ 4x + 10 = 0 \implies x = -\frac{10}{4} = -\frac{5}{2} \]
The x-intercept is \( (-\frac{5}{2}, 0) \).
- To find the y-intercept, set \( x = 0 \):
\[ b_2y + 10 = 0 \implies y = -\frac{10}{b_2} \]
The y-intercept is \( (0, -\frac{10}{b_2}) \).

3. **Concyclic Condition**:
The four points \( (-\frac{5}{3}, 0) \), \( (0, -\frac{5}{b_1}) \), \( (-\frac{5}{2}, 0) \), and \( (0, -\frac{10}{b_2}) \) are concyclic if the following condition holds:
\[ a_1 \cdot a_2 = b_1 \cdot b_2 \]
where \( a_1 = 3 \) (from Equation 1), \( a_2 = 4 \) (from Equation 2).

4. **Substituting Values**:
From the condition \( a_1 \cdot a_2 = b_1 \cdot b_2 \):
\[ 3 \cdot 4 = b_1 \cdot b_2 \]
\[ 12 = b_1 \cdot b_2 \]

5. **Conclusion**:
Therefore, the value of \( b_1 b_2 \) is \( 12 \).

### Final Answer:
\[ b_1 b_2 = 12 \]