Consider an ellipse E:`x^(2)/a^(2)+y^(2)/b^(2)=1`, centered at point 'O' and having AB and CD as its major and minor axes respectively if `S_1`be one of the focus of the ellipse, radius of the incircle of `∆OCS_1` be 1 unit and `OS_1=6 `units. Q. The equation of the director circle of (E) is

Options

A. (a)`x^(2)+y^(2)=48.5`
B. (b)`x^(2)+y^(2)=97`
C. (c)`x^(2)+y^(2)=sqrt48.5`
D. (d)`x^(2)+y^(2)=sqrt97`

Answer

To solve the problem, we need to find the equation of the director circle of the ellipse given by the equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 1: Understand the parameters of the ellipse
The foci of the ellipse are located at a distance \(c\) from the center, where \(c = \sqrt{a^2 - b^2}\). We are given that \(OS_1 = 6\) units, which means \(c = 6\).

### Step 2: Relate \(a\), \(b\), and \(c\)
From the relationship between \(a\), \(b\), and \(c\), we have:

\[ c^2 = a^2 - b^2 \]
Substituting \(c = 6\):

\[ 36 = a^2 - b^2 \quad \text{(1)} \]

### Step 3: Use the radius of the incircle of triangle \(OCS_1\)
The radius \(r\) of the incircle of triangle \(OCS_1\) is given as 1 unit. The area \(A\) of triangle \(OCS_1\) can be expressed in terms of its semi-perimeter \(s\) and the radius \(r\):

\[ r = \frac{A}{s} \]

The semi-perimeter \(s\) of triangle \(OCS_1\) is given by:

\[ s = \frac{OC + CS_1 + OS_1}{2} = \frac{b + a + 6}{2} \]

### Step 4: Find the area \(A\) of triangle \(OCS_1\)
The area \(A\) can also be calculated using the formula:

\[ A = \frac{1}{2} \times OC \times OS_1 = \frac{1}{2} \times b \times 6 = 3b \]

### Step 5: Set up the equation using the incircle radius
Substituting the area and semi-perimeter into the radius formula gives:

\[ 1 = \frac{3b}{\frac{b + a + 6}{2}} \]

Cross-multiplying yields:

\[ b + a + 6 = 6b \]
\[ 5b = a + 6 \quad \text{(2)} \]

### Step 6: Substitute equation (2) into equation (1)
From equation (2), we can express \(a\) in terms of \(b\):

\[ a = 5b - 6 \]

Substituting this into equation (1):

\[ 36 = (5b - 6)^2 - b^2 \]

Expanding and simplifying:

\[ 36 = 25b^2 - 60b + 36 - b^2 \]
\[ 0 = 24b^2 - 60b \]
\[ 0 = 12b(2b - 5) \]

Thus, \(b = 0\) or \(b = \frac{5}{2}\). Since \(b\) cannot be zero, we have:

\[ b = \frac{5}{2} \]

### Step 7: Find \(a\)
Substituting \(b\) back into equation (2):

\[ a = 5 \cdot \frac{5}{2} - 6 = \frac{25}{2} - 6 = \frac{25}{2} - \frac{12}{2} = \frac{13}{2} \]

### Step 8: Calculate \(a^2 + b^2\)
Now we can calculate \(a^2\) and \(b^2\):

\[ a^2 = \left(\frac{13}{2}\right)^2 = \frac{169}{4} \]
\[ b^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \]

Adding these gives:

\[ a^2 + b^2 = \frac{169}{4} + \frac{25}{4} = \frac{194}{4} = \frac{97}{2} \]

### Step 9: Write the equation of the director circle
The equation of the director circle of the ellipse is given by:

\[ x^2 + y^2 = a^2 + b^2 \]

Thus, the equation is:

\[ x^2 + y^2 = \frac{97}{2} \]

### Final Answer
The equation of the director circle of the ellipse is:

\[ \boxed{x^2 + y^2 = \frac{97}{2}} \]