A square OABC is formed by line pairs xy = 0 and xy+1= x+y where 'O' is the origin. A circle with centre C1 inside the square is drawn to touch the line pair xy=0 and another circle with centre C2 and radius twice that of C1, is drawn to touch the circle C1 and the other line pair. The radius of the circle with centre C1 is:

Radius of Circle C1

A square OABC is formed by the lines:

• xy = 0 (x-axis and y-axis)
• xy + 1 = x + y, which simplifies to (x - 1)(y - 1) = 0, representing the lines x = 1 and y = 1.

The vertices of the square are (0,0), (1,0), (0,1), and (1,1), with side length 1.

A circle C1 is drawn inside the square, touching the x-axis and y-axis, and another circle C2 has a radius twice that of C1. Circle C2 touches C1, as well as the lines x = 1 and y = 1. We need to find the radius of C1.

Properties of Circle C1

The circle C1 has its center at (r1, r1), where r1 is the radius. Since the circle touches both the x-axis and y-axis, this is straightforward.

Properties of Circle C2

The radius of C2 is r2 = 2r1. The center of C2 is at (1 - r2, 1 - r2) because it touches the lines x = 1 and y = 1.

The distance between the centers of C1 and C2 is equal to the sum of their radii:

Distance between centers = r1 + r2

Using the Euclidean distance formula:

√[(1 - r2 - r1)² + (1 - r2 - r1)²] = r1 + r2

Squaring both sides:

2(1 - r2 - r1)² = (r1 + r2)²

Substitute r2 = 2r1 into the equation:

2(1 - 2r1 - r1)² = (r1 + 2r1)²

Simplify:

2(1 - 3r1)² = (3r1)²

Expanding both sides:

2(1 - 6r1 + 9r1²) = 9r1²

Simplify further:

2 - 12r1 + 18r1² = 9r1²

2 - 12r1 + 9r1² = 0

Using the quadratic formula:

r1 = [-b ± √(b² - 4ac)] / 2a

Here, a = 9, b = -12, c = 2. Substituting:

r1 = [12 ± √(144 - 72)] / 18

r1 = [12 ± √72] / 18

Simplify √72 = 6√2:

r1 = [12 ± 6√2] / 18

r1 = 2 ± √2 / 3

Since r1 must be positive:

r1 = (2 - √2) / 3

Final Answer

The radius of the circle C1 is:

r1 = (√2) / [3(√2 + 1)]