If the circles `x^(2)+y^(2) +5Kx+2y + K=0 and2(x^(2)+y^(2))+2Kx +3y-1=0, (KinR)`, intersect at the points P and Q, then the line 4x + 5y - K = 0 passes through P and Q, for

Options

A. no value of K
B. exactly one value of K
C. exactly two value of K
D. infinitely many value of K

Explanation

To solve the problem step by step, we need to analyze the given circles and the line.

### Step 1: Write down the equations of the circles
The equations of the circles are given as:
1. \( x^2 + y^2 + 5Kx + 2y + K = 0 \) (Circle 1)
2. \( 2(x^2 + y^2) + 2Kx + 3y - 1 = 0 \) (Circle 2)

### Step 2: Simplify the second circle's equation
We can simplify the second circle's equation by dividing everything by 2:
\[ x^2 + y^2 + Kx + \frac{3}{2}y - \frac{1}{2} = 0 \]
This is now the equation of Circle 2.

### Step 3: Set up the equations for the common chord
The common chord of two circles can be found using the formula \( S_1 - S_2 = 0 \), where \( S_1 \) and \( S_2 \) are the equations of the circles.

Let:
- \( S_1 = x^2 + y^2 + 5Kx + 2y + K \)
- \( S_2 = x^2 + y^2 + Kx + \frac{3}{2}y - \frac{1}{2} \)

Now, we find \( S_1 - S_2 \):
\[ S_1 - S_2 = (x^2 + y^2 + 5Kx + 2y + K) - (x^2 + y^2 + Kx + \frac{3}{2}y - \frac{1}{2}) = 0 \]
This simplifies to:
\[ (5K - K)x + \left(2 - \frac{3}{2}\right)y + \left(K + \frac{1}{2}\right) = 0 \]
\[ (4K)x + \frac{1}{2}y + \left(K + \frac{1}{2}\right) = 0 \]

### Step 4: Rearranging the equation
We can rearrange this equation to get:
\[ 4Kx + \frac{1}{2}y + K + \frac{1}{2} = 0 \]

### Step 5: Compare with the line equation
The line equation given is:
\[ 4x + 5y - K = 0 \]
We need to ensure that the coefficients of \( x \) and \( y \) in the common chord equation match those in the line equation.

### Step 6: Set up the equations based on coefficients
From the equations:
1. Coefficient of \( x \): \( 4K = 4 \) → \( K = 1 \)
2. Coefficient of \( y \): \( \frac{1}{2} = 5 \) → This is not possible.

### Step 7: Solve for K from the first coefficient
From the first coefficient:
\[ 4K = 4 \implies K = 1 \]

### Step 8: Check for contradictions
Now substituting \( K = 1 \) into the second coefficient gives:
\[ \frac{1}{2} = 5 \implies \text{Contradiction} \]

### Conclusion
Since we have reached a contradiction, it means there is no value of \( K \) such that the line \( 4x + 5y - K = 0 \) passes through both points \( P \) and \( Q \).

### Final Answer
There is no value of \( K \) for which the line passes through points \( P \) and \( Q \).

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