If the lines `a_(1)x+b_(1)y+c_(1)=0` and `a_(2)x+b_(2)y+c_(2)=0` cut the coordinate axes in concyclic points, then

Options

A. `a_(1) a_(2)=b_(1)b_(2)`
B. `a_(1)b_(1)=a_(2)b_(2)`
C. `a_(1)//a_(2)=b_(1)//b_(2)`
D. none

Answer

To solve the problem, we need to analyze the conditions under which the intersection points of the given lines with the coordinate axes are concyclic. Let's break this down step by step.

### Step 1: Write the equations of the lines
We have two lines given by the equations:
1. \( a_1 x + b_1 y + c_1 = 0 \)
2. \( a_2 x + b_2 y + c_2 = 0 \)

### Step 2: Convert the lines to intercept form
To find the intercepts of these lines on the coordinate axes, we convert them to intercept form.

For the first line:
\[ a_1 x + b_1 y = -c_1 \]
Dividing by \(-c_1\):
\[ \frac{x}{-\frac{c_1}{a_1}} + \frac{y}{-\frac{c_1}{b_1}} = 1 \]
This gives us the x-intercept \( A \left(-\frac{c_1}{a_1}, 0\right) \) and the y-intercept \( B \left(0, -\frac{c_1}{b_1}\right) \).

For the second line:
\[ a_2 x + b_2 y = -c_2 \]
Dividing by \(-c_2\):
\[ \frac{x}{-\frac{c_2}{a_2}} + \frac{y}{-\frac{c_2}{b_2}} = 1 \]
This gives us the x-intercept \( C \left(-\frac{c_2}{a_2}, 0\right) \) and the y-intercept \( D \left(0, -\frac{c_2}{b_2}\right) \).

### Step 3: Establish the condition for concyclic points
The points \( A, B, C, D \) are concyclic if the following condition holds:
\[ OA \cdot OC = OB \cdot OD \]
where \( O \) is the origin.

### Step 4: Calculate distances
The distances from the origin to the points are:
- \( OA = \left| -\frac{c_1}{a_1} \right| = \frac{|c_1|}{|a_1|} \)
- \( OB = \left| -\frac{c_1}{b_1} \right| = \frac{|c_1|}{|b_1|} \)
- \( OC = \left| -\frac{c_2}{a_2} \right| = \frac{|c_2|}{|a_2|} \)
- \( OD = \left| -\frac{c_2}{b_2} \right| = \frac{|c_2|}{|b_2|} \)

### Step 5: Substitute into the concyclic condition
Substituting these distances into the concyclic condition:
\[ \frac{|c_1|}{|a_1|} \cdot \frac{|c_2|}{|a_2|} = \frac{|c_1|}{|b_1|} \cdot \frac{|c_2|}{|b_2|} \]

### Step 6: Simplify the equation
Cancelling \( |c_1| \) and \( |c_2| \) from both sides (assuming they are non-zero), we get:
\[ \frac{1}{|a_1|} \cdot \frac{1}{|a_2|} = \frac{1}{|b_1|} \cdot \frac{1}{|b_2|} \]
This simplifies to:
\[ a_1 a_2 = b_1 b_2 \]

### Conclusion
Thus, we conclude that if the lines cut the coordinate axes in concyclic points, then:
\[ a_1 a_2 = b_1 b_2 \]