Consider the two circles `C_(1):x^(2)+y^(2)=a^(2)andC_(2):x^(2)+y^(2)=b^(2)(agtb)` Let A be a fixed point on the circle `C_(1)`, say A(a,0) and B be a variable point on the circle `C_(2)`. The line BA meets the circle `C_(2)` again at C. 'O' being the origin.
If `(OA)^(2)+(OB)^(2)+(BC)^(2)=lamda," then "lamdain`

Options

A. (a)`(b^(2)+a^(2),5b^(2)+a^(2)]`
B. (b)`[4b^(2),4b^(2)+a^(2)]`
C. (c)`[4a^(2),4b^(2)]`
D. (d)`[5b^(2)-3a^(2),5b^(2)+3a^(2)]`

Answer

To solve the problem, we need to analyze the given circles and the points involved.

### Step 1: Understand the Circles and Points
We have two circles:
- Circle \( C_1: x^2 + y^2 = a^2 \) (with center at the origin and radius \( a \))
- Circle \( C_2: x^2 + y^2 = b^2 \) (with center at the origin and radius \( b \), where \( a > b \))

Let \( A \) be the fixed point on circle \( C_1 \) at coordinates \( (a, 0) \). Let \( B \) be a variable point on circle \( C_2 \) with coordinates \( (b \cos \theta, b \sin \theta) \), where \( \theta \) is a variable angle.

### Step 2: Find the Coordinates of Point B
The coordinates of point \( B \) on circle \( C_2 \) can be expressed as:
\[ B(b \cos \theta, b \sin \theta) \]

### Step 3: Calculate \( OA^2 \) and \( OB^2 \)
The distance from the origin \( O \) to point \( A \) is:
\[ OA^2 = a^2 \]
The distance from the origin \( O \) to point \( B \) is:
\[ OB^2 = b^2 \]

### Step 4: Determine the Equation of Line \( BA \)
The line \( BA \) can be expressed in parametric form. The slope of line \( BA \) from \( A(a, 0) \) to \( B(b \cos \theta, b \sin \theta) \) is:
\[ \text{slope} = \frac{b \sin \theta - 0}{b \cos \theta - a} \]
The equation of line \( BA \) can be written as:
\[ y - 0 = \frac{b \sin \theta}{b \cos \theta - a} (x - a) \]

### Step 5: Find Intersection Point \( C \)
To find point \( C \), we substitute the equation of line \( BA \) into the equation of circle \( C_2 \):
\[ x^2 + y^2 = b^2 \]
This will yield a quadratic equation in \( x \) after substituting \( y \) from the line equation.

### Step 6: Calculate \( BC^2 \)
The length \( BC \) can be calculated using the coordinates of points \( B \) and \( C \). If \( C \) is the second intersection point, we can find \( BC^2 \) as:
\[ BC^2 = (x_C - x_B)^2 + (y_C - y_B)^2 \]

### Step 7: Combine the Results
We need to find:
\[ OA^2 + OB^2 + BC^2 = \lambda \]
Substituting the values we have:
\[ \lambda = a^2 + b^2 + BC^2 \]

### Step 8: Determine Maximum and Minimum Values
As \( B \) moves around circle \( C_2 \), \( BC \) will vary. The maximum value of \( BC^2 \) occurs when \( B \) is at its farthest point from \( A \) and the minimum occurs when \( B \) is closest to \( A \).

### Final Expression
Thus, we can conclude:
\[ \lambda_{max} = a^2 + b^2 + (2b)^2 = a^2 + b^2 + 4b^2 = a^2 + 5b^2 \]
\[ \lambda_{min} = a^2 + b^2 + 0 = a^2 + b^2 \]

### Conclusion
The values of \( \lambda \) can vary between \( a^2 + b^2 \) and \( a^2 + 5b^2 \).