The values of `lamda` for which the circle `x^(2)+y^(2)+6x+5+lamda(x^(2)+y^(2)-8x+7)=0` dwindles into a point are

Options

A. `1pm(sqrt2)/(3)`
B. `2pm(2sqrt2)/(3)`
C. `2pm(4sqrt2)/(3)`
D. `1pm(4sqrt2)/(3)`

Answer

To find the values of \( \lambda \) for which the given circle equation dwindles into a point, we will follow these steps:

### Step 1: Write the given equation
The equation of the circle is given as:
\[ x^2 + y^2 + 6x + 5 + \lambda(x^2 + y^2 - 8x + 7) = 0 \]

### Step 2: Expand the equation
Expanding the equation, we have:
\[ x^2 + y^2 + 6x + 5 + \lambda x^2 + \lambda y^2 - 8\lambda x + 7\lambda = 0 \]
Combining like terms, we get:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + (6 - 8\lambda)x + (5 + 7\lambda) = 0 \]

### Step 3: Identify the coefficients
From the standard form of the circle equation \( Ax^2 + Ay^2 + 2Gx + 2Fy + C = 0 \), we can identify:
- \( A = 1 + \lambda \)
- \( 2G = 6 - 8\lambda \) → \( G = \frac{6 - 8\lambda}{2} = 3 - 4\lambda \)
- \( 2F = 0 \) → \( F = 0 \)
- \( C = 5 + 7\lambda \)

### Step 4: Condition for the circle to be a point
For the circle to dwindle into a point, the radius must be zero. The radius \( r \) is given by:
\[ r = \sqrt{G^2 + F^2 - C} \]
Substituting the values we found:
\[ r = \sqrt{(3 - 4\lambda)^2 + 0^2 - (5 + 7\lambda)} \]
Setting \( r = 0 \):
\[ (3 - 4\lambda)^2 - (5 + 7\lambda) = 0 \]

### Step 5: Expand and simplify
Expanding the equation:
\[ (3 - 4\lambda)^2 = 5 + 7\lambda \]
\[ 9 - 24\lambda + 16\lambda^2 = 5 + 7\lambda \]
Rearranging gives:
\[ 16\lambda^2 - 31\lambda + 4 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 16 \), \( b = -31 \), and \( c = 4 \):
\[ \lambda = \frac{31 \pm \sqrt{(-31)^2 - 4 \cdot 16 \cdot 4}}{2 \cdot 16} \]
Calculating the discriminant:
\[ \lambda = \frac{31 \pm \sqrt{961 - 256}}{32} \]
\[ \lambda = \frac{31 \pm \sqrt{705}}{32} \]

### Step 7: Simplify the square root
The square root can be simplified:
\[ \sqrt{705} = \sqrt{(25)(28.2)} \approx 26.5 \]
Thus, we find:
\[ \lambda = \frac{31 \pm 4\sqrt{2}}{32} \]

### Final Answer
The values of \( \lambda \) for which the circle dwindles into a point are:
\[ \lambda = 2 \pm \frac{4\sqrt{2}}{3} \]