Two conics `a_1x^2+2h_1xy + b_1y^2 = c_1, a_2x^2 + 2h_2xy+b_2y^2 = c_2` intersect in 4 concyclic points. Then

Options

A. `(a_(1)-b_(1)h_(2)=(a_(2)-b_(2))h_(1)`
B. `(A_(1)-b_(1))h_(1)=(a_(2)-b_(2))h_(2)`
C. `(a_(1)+b_(1))h_(2)=(a_(2)+b_(2))h_(1)`
D. `(a_(1)+b_(1))h_(1)=(a_(2)+b_(2))h_(2)`

Explanation

To solve the problem, we need to find the relationship between the coefficients of the two conics given that they intersect in four concyclic points.

### Step-by-Step Solution:

1. **Write the equations of the conics:**
The two conics are given as:
\[ S_1: a_1 x^2 + 2h_1 xy + b_1 y^2 = c_1 \]
\[ S_2: a_2 x^2 + 2h_2 xy + b_2 y^2 = c_2 \]

2. **Form the combined equation:**
Since the conics intersect at four points, we can express their intersection in the form:
\[ S_1 + \lambda S_2 = 0 \]
This leads to:
\[ (a_1 + \lambda a_2)x^2 + (2h_1 + \lambda 2h_2)xy + (b_1 + \lambda b_2)y^2 - (c_1 + \lambda c_2) = 0 \]

3. **Conditions for concyclic points:**
For the points of intersection to be concyclic, the following conditions must hold:
- The coefficient of \(x^2\) must equal the coefficient of \(y^2\):
\[ a_1 + \lambda a_2 = b_1 + \lambda b_2 \]
- The coefficient of \(xy\) must equal zero:
\[ 2h_1 + \lambda 2h_2 = 0 \]

4. **Solve for \(\lambda\):**
From the second condition:
\[ 2h_1 + \lambda 2h_2 = 0 \implies \lambda = -\frac{h_1}{h_2} \]

5. **Substitute \(\lambda\) into the first condition:**
Substitute \(\lambda\) into the first condition:
\[ a_1 - \frac{h_1}{h_2} a_2 = b_1 - \frac{h_1}{h_2} b_2 \]
Rearranging gives:
\[ a_1 - b_1 = \frac{h_1}{h_2}(a_2 - b_2) \]

6. **Final relationship:**
We can rewrite this as:
\[ (a_2 - b_2)h_1 = (a_1 - b_1)h_2 \]

### Conclusion:
The relationship derived from the conditions for the conics intersecting at four concyclic points is:
\[ (a_2 - b_2)h_1 = (a_1 - b_1)h_2 \]