IF `x^(2) + y^(2) - 2y - 15 + lambda (2x + y - 9) = 0` represents family of circles for. Different values of `lambda`, then the equation of the circle(s) along these circles having minimum radius is/are

Options

A. `3x^(2) + 3y^(2) -2x - 7y - 36 = 0`
B. `x^(2) + y^(2) - 2y - 15 = 0`
C. `5x^(2) + 5y^(2) - 32x - 26y + 69 = 0`
D. `x^(2) + y^(2) -10x - 7y + 30 = 0`

Answer

To solve the problem, we need to analyze the given equation of the family of circles and find the one with the minimum radius. Let's break it down step by step.

### Step 1: Rewrite the given equation
The given equation is:
\[ x^2 + y^2 - 2y - 15 + \lambda(2x + y - 9) = 0 \]

We can rearrange this equation:
\[ x^2 + y^2 + 2\lambda x + (-2 + \lambda)y + (-9\lambda - 15) = 0 \]

### Step 2: Identify coefficients
This equation can be compared to the standard form of a circle:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From this comparison, we can identify:
- \( 2g = 2\lambda \) → \( g = \lambda \)
- \( 2f = -2 + \lambda \) → \( f = \frac{\lambda - 2}{2} \)
- \( c = -9\lambda - 15 \)

### Step 3: Find the radius
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 \( c \):
\[ R = \sqrt{\lambda^2 + \left(\frac{\lambda - 2}{2}\right)^2 - (-9\lambda - 15)} \]

### Step 4: Simplify the radius expression
Now, let's simplify the expression for \( R^2 \):
\[ R^2 = \lambda^2 + \frac{(\lambda - 2)^2}{4} + 9\lambda + 15 \]
\[ = \lambda^2 + \frac{\lambda^2 - 4\lambda + 4}{4} + 9\lambda + 15 \]
\[ = \lambda^2 + \frac{1}{4}(\lambda^2 - 4\lambda + 4) + 9\lambda + 15 \]
\[ = \lambda^2 + \frac{1}{4}\lambda^2 - \lambda + 1 + 9\lambda + 15 \]
\[ = \frac{5}{4}\lambda^2 + 8\lambda + 16 \]

### Step 5: Differentiate to find minimum radius
To find the minimum radius, we need to differentiate \( R^2 \) with respect to \( \lambda \):
\[ \frac{d(R^2)}{d\lambda} = \frac{5}{2}\lambda + 8 \]
Setting the derivative to zero to find critical points:
\[ \frac{5}{2}\lambda + 8 = 0 \]
\[ \lambda = -\frac{16}{5} \]

### Step 6: Substitute \( \lambda \) back to find the equation of the circle
Now substituting \( \lambda = -\frac{16}{5} \) back into the equation of the circle:
\[ x^2 + y^2 + 2\left(-\frac{16}{5}\right)x + \left(-2 - \frac{16}{5}\right)y + \left(-9\left(-\frac{16}{5}\right) - 15\right) = 0 \]
Calculating each term:
- \( g = -\frac{16}{5} \)
- \( f = -\frac{2 - 16/5}{2} = -\frac{-6/5}{2} = \frac{3}{5} \)
- \( c = \frac{144}{5} - 15 = \frac{144 - 75}{5} = \frac{69}{5} \)

Thus, the equation becomes:
\[ x^2 + y^2 - \frac{32}{5}x + \frac{3}{5}y + \frac{69}{5} = 0 \]
Multiplying through by 5 to eliminate the fractions:
\[ 5x^2 + 5y^2 - 32x + 3y + 69 = 0 \]

### Final Answer
The equation of the circle with minimum radius is:
\[ 5x^2 + 5y^2 - 32x + 3y + 69 = 0 \]