The two conics `ax^(2)+2hxy+by^(2)=c` and `px^(2)+2kxy+qy^(2)=r` intersect in four concyclic points, then

Options

A. `(a-b)/(p-q)=(h)/(k)`
B. `(a-b)/(p-q)=(k)/(h)`
C. `(a+b)/(p+q)=(h)/(k)`
D. `(a+b)/(p+q)=(k)/(h)`

Answer

To solve the problem, we need to establish the condition under which the two conics intersect at four concyclic points. Let's break down the solution step by step.

### Step 1: Write the equations of the conics
We have two conics given by:
1. \( ax^2 + 2hxy + by^2 = c \) (Equation 1)
2. \( px^2 + 2kxy + qy^2 = r \) (Equation 2)

### Step 2: Multiply the equations
To facilitate the elimination of variables, we multiply the first equation by \( k \) and the second equation by \( h \):
1. \( akx^2 + 2hkxy + bky^2 = ck \) (Equation 3)
2. \( phx^2 + 2hky + qhy^2 = rh \) (Equation 4)

### Step 3: Subtract the two equations
Next, we subtract Equation 4 from Equation 3:
\[ (ak - ph)x^2 + (2hk - 2hk)xy + (bk - qh)y^2 = ck - rh \]
This simplifies to:
\[ (ak - ph)x^2 + (bk - qh)y^2 = ck - rh \]

### Step 4: Condition for concyclic points
For the two conics to intersect at four concyclic points, the resulting equation from the subtraction must represent a circle. A necessary condition for this is that the coefficients of \( x^2 \) and \( y^2 \) must be equal, and the coefficient of \( xy \) must be zero.

Thus, we set up the following conditions:
1. Coefficient of \( x^2 \) must equal the coefficient of \( y^2 \):
\[ ak - ph = bk - qh \]
Rearranging gives:
\[ ak - bk = ph - qh \]
or
\[ (a - b)k = (p - q)h \]

### Step 5: Final condition
From the above equation, we can derive the final condition:
\[ \frac{a - b}{p - q} = \frac{h}{k} \]

### Conclusion
Thus, the condition for the two conics to intersect at four concyclic points is:
\[ \frac{a - b}{p - q} = \frac{h}{k} \]