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

Options

A. `(3pm 4sqrt(2))/(3)`
B. `(6pm 4sqrt(2))/(3)`
C. `(4pm 3sqrt(2))/(3)`
D. none

Answer

To find the values of \( \lambda \) for which the given circle equation dwindles into a point, we start with the equation:

\[ x^2 + y^2 + 6x + 5 + \lambda(x^2 + y^2 - 8x + 7) = 0 \]

### Step 1: Combine the terms

We can rewrite the equation by combining like terms:

\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + (6 - 8\lambda)x + (5 + 7\lambda) = 0 \]

### Step 2: Identify coefficients

From the equation, we can identify the coefficients:
- Coefficient of \( x^2 \) and \( y^2 \): \( 1 + \lambda \)
- Coefficient of \( x \): \( 6 - 8\lambda \)
- Constant term: \( 5 + 7\lambda \)

### Step 3: Standard form of a circle

The standard form of a circle is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Where \( g = -h \), \( f = -k \), and \( c = r^2 \). For our equation, we can express it in the form:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

### Step 4: Calculate \( g \), \( f \), and \( c \)

From our coefficients:
- \( g = \frac{6 - 8\lambda}{2} = 3 - 4\lambda \)
- \( f = 0 \) (since there is no \( y \) term)
- \( c = 5 + 7\lambda \)

### Step 5: 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:

\[ r^2 = g^2 + f^2 - c \]

Setting \( r^2 = 0 \):

\[ (3 - 4\lambda)^2 + 0^2 - (5 + 7\lambda) = 0 \]

### Step 6: Solve the equation

Expanding the equation:

\[ (3 - 4\lambda)^2 - (5 + 7\lambda) = 0 \]

Expanding \( (3 - 4\lambda)^2 \):

\[ 9 - 24\lambda + 16\lambda^2 - 5 - 7\lambda = 0 \]

Combining like terms:

\[ 16\lambda^2 - 31\lambda + 4 = 0 \]

### Step 7: Use the quadratic formula

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:

\[ \lambda = \frac{31 \pm \sqrt{961 - 256}}{32} \]
\[ \lambda = \frac{31 \pm \sqrt{705}}{32} \]

### Final Step: Simplify the result

Thus, the values of \( \lambda \) for which the circle dwindles into a point are:

\[ \lambda = \frac{31 \pm \sqrt{705}}{32} \]