The number of integral values of `lambda` for which the equation `x^(2)+y^(2)-2lambdax+2lambday+14=0` represents a circle whose radius cannot exceed 6, is

Options

A. 10
B. 11
C. 12
D. 9

Explanation

To solve the problem, we need to analyze the given equation of the circle and find the integral values of \( \lambda \) for which the radius of the circle does not exceed 6.

### Step 1: Write the given equation
The given equation is:
\[ x^2 + y^2 - 2\lambda x + 2\lambda y + 14 = 0 \]

### Step 2: Compare with the standard form of a circle
The standard form of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the comparison, we can identify:
- \( 2g = -2\lambda \) → \( g = -\lambda \)
- \( 2f = 2\lambda \) → \( f = \lambda \)
- \( c = 14 \)

### Step 3: Find the radius of the circle
The radius \( r \) of the circle can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values of \( g \), \( f \), and \( c \):
\[ r = \sqrt{(-\lambda)^2 + (\lambda)^2 - 14} \]
This simplifies to:
\[ r = \sqrt{\lambda^2 + \lambda^2 - 14} = \sqrt{2\lambda^2 - 14} \]

### Step 4: Set the condition for the radius
We need the radius \( r \) to not exceed 6:
\[ \sqrt{2\lambda^2 - 14} \leq 6 \]

### Step 5: Square both sides
Squaring both sides to eliminate the square root gives:
\[ 2\lambda^2 - 14 \leq 36 \]

### Step 6: Rearrange the inequality
Rearranging the inequality:
\[ 2\lambda^2 \leq 50 \]
\[ \lambda^2 \leq 25 \]

### Step 7: Determine the range of \( \lambda \)
Taking the square root of both sides:
\[ -\sqrt{25} \leq \lambda \leq \sqrt{25} \]
This simplifies to:
\[ -5 \leq \lambda \leq 5 \]

### Step 8: Count the integral values of \( \lambda \)
The integral values of \( \lambda \) in the range from -5 to 5 are:
\[ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 \]
Counting these values gives us a total of 11 integral values.

### Final Answer
The number of integral values of \( \lambda \) for which the equation represents a circle whose radius cannot exceed 6 is:
\[ \boxed{11} \]