If the lines `a_(1)x+b_(1)y+c_(1)=0` and `a_(2)x+b_(2)y+c_(2)=0` cut the co-ordinate axes in concylic points, then

Options

A. `a_(1)b_(1)=a_(2)b_(2)`
B. `a_(1)//a_(2)=b_(1)//b_(2)`
C. `a_(1)+a_(2)=b_(1)+b_(2)`
D. `a_(1)a_(2)=b_(1)b_(2)`

Answer

To solve the problem, we need to show that if the lines \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) cut the coordinate axes at concyclic points, then the relationship \( a_1 a_2 = b_1 b_2 \) holds.

### Step-by-Step Solution:

1. **Convert the lines to intercept form:**
- For the first line \( L_1: a_1x + b_1y + c_1 = 0 \):
\[ \frac{x}{-\frac{c_1}{a_1}} + \frac{y}{-\frac{c_1}{b_1}} = 1 \]
This means the x-intercept is \( A\left(-\frac{c_1}{a_1}, 0\right) \) and the y-intercept is \( B\left(0, -\frac{c_1}{b_1}\right) \).

- For the second line \( L_2: a_2x + b_2y + c_2 = 0 \):
\[ \frac{x}{-\frac{c_2}{a_2}} + \frac{y}{-\frac{c_2}{b_2}} = 1 \]
This means the x-intercept is \( C\left(-\frac{c_2}{a_2}, 0\right) \) and the y-intercept is \( D\left(0, -\frac{c_2}{b_2}\right) \).

2. **Identify the points:**
- The points of intersection with the axes are:
- \( A = \left(-\frac{c_1}{a_1}, 0\right) \)
- \( B = \left(0, -\frac{c_1}{b_1}\right) \)
- \( C = \left(-\frac{c_2}{a_2}, 0\right) \)
- \( D = \left(0, -\frac{c_2}{b_2}\right) \)

3. **Use the condition for concyclic points:**
- The points \( A, B, C, D \) are concyclic if the product of the lengths of the segments from the origin to the points satisfies:
\[ OA \cdot OC = OB \cdot OD \]
- Where:
- \( OA = -\frac{c_1}{a_1} \)
- \( OB = -\frac{c_1}{b_1} \)
- \( OC = -\frac{c_2}{a_2} \)
- \( OD = -\frac{c_2}{b_2} \)

4. **Substituting the lengths into the condition:**
\[ \left(-\frac{c_1}{a_1}\right) \cdot \left(-\frac{c_2}{a_2}\right) = \left(-\frac{c_1}{b_1}\right) \cdot \left(-\frac{c_2}{b_2}\right) \]
This simplifies to:
\[ \frac{c_1 c_2}{a_1 a_2} = \frac{c_1 c_2}{b_1 b_2} \]

5. **Cancelling \( c_1 c_2 \) (assuming they are non-zero):**
\[ \frac{1}{a_1 a_2} = \frac{1}{b_1 b_2} \]
This leads to:
\[ a_1 a_2 = b_1 b_2 \]

### Conclusion:
Thus, we have shown that if the lines cut the coordinate axes at concyclic points, then the relationship \( a_1 a_2 = b_1 b_2 \) holds true.