A square OABC is formed by line pairs `xy=0` and `xy+1=x+y` where o is the origin . A circle with circle `C_1` inside the square is drawn to touch the line pair `xy=0` and another circle with centre `C_2` and radius twice that `C_1` is drawn touch the circle `C_1` and the other line .The radius of the circle with centre `C_1`

Options

A. `(sqrt(2))/(sqrt(3)(sqrt(2)+1))`
B. `(2sqrt(2))/(3(sqrt(2)+1))`
C. `(sqrt(2))/(3(sqrt(2)+1))`
D. `(sqrt(2)+1)/(3sqrt(2))`

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript and derive the radius of the circle with center \( C_1 \).

### Step 1: Identify the square OABC
The square OABC is formed by the line pairs \( xy = 0 \) (which represents the x-axis and y-axis) and \( xy + 1 = x + y \).

1. **Find the points of intersection:**
- For \( xy = 0 \): The points are \( O(0, 0) \), \( A(1, 0) \), and \( B(0, 1) \).
- For \( xy + 1 = x + y \): Rearranging gives \( xy - x - y + 1 = 0 \). This can be factored to find the points of intersection.

### Step 2: Solve the equation \( xy + 1 = x + y \)
Rearranging gives:
\[ xy - x - y + 1 = 0 \implies (x - 1)(y - 1) = 0 \]
This gives the points \( (1, 0) \) and \( (0, 1) \), confirming that the square has vertices at \( O(0, 0) \), \( A(1, 0) \), \( B(0, 1) \), and \( C(1, 1) \).

### Step 3: Define the circles
Let the radius of circle \( C_1 \) be \( r \). The circle \( C_1 \) is inscribed in the square and touches the x-axis and y-axis.

### Step 4: Radius of circle \( C_1 \)
Since the circle \( C_1 \) touches both axes, its radius \( r \) can be defined as:
\[ r = 1 - r \]
This leads to:
\[ 1 = 2r \implies r = \frac{1}{2} \]

### Step 5: Define circle \( C_2 \)
Circle \( C_2 \) has a radius of \( 2r \):
\[ \text{Radius of } C_2 = 2r = 2 \times \frac{1}{2} = 1 \]

### Step 6: Distance between centers \( C_1 \) and \( C_2 \)
The distance between the centers of circles \( C_1 \) and \( C_2 \) is given by:
\[ \text{Distance} = r + 2r = 3r \]

### Step 7: Set up the equation for the remaining distance
The remaining distance from the point where circle \( C_2 \) touches the line \( y = 1 \) is:
\[ 1 - 2r \quad \text{(for circle } C_2\text{)} \]
And for circle \( C_1 \):
\[ 1 - r \]

### Step 8: Solve the quadratic equation
Setting up the equation:
\[ (1 - 2r) + (1 - r) = 3r \]
This simplifies to:
\[ 2 - 3r = 3r \implies 6r = 2 \implies r = \frac{1}{3} \]

### Step 9: Conclusion
Thus, the radius of circle \( C_1 \) is:
\[ \boxed{\frac{1}{3}} \]