inside the circles `x^2+y^2=1` there are three circles of equal radius `a` tangent to each other and to `s` the value of `a` equals to

Options

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

Explanation

To solve the problem of finding the radius \( a \) of three equal circles tangent to each other and to the larger circle defined by the equation \( x^2 + y^2 = 1 \), we can follow these steps:

### Step 1: Understand the Geometry
The larger circle has a radius of 1 (since \( x^2 + y^2 = 1 \)). The centers of the three smaller circles (each with radius \( a \)) will form an equilateral triangle, and each circle will be tangent to the larger circle and to each other.

### Step 2: Determine the Distance Between Centers
The distance between the centers of any two smaller circles (say \( A \) and \( C \)) is equal to the sum of their radii:
\[ AC = a + a = 2a \]

### Step 3: Relate the Centers to the Larger Circle
The distance from the center of the larger circle (point \( O \)) to the center of any smaller circle (point \( A \) or \( C \)) is:
\[ OA = OC = 1 - a \]
This is because the distance from the center of the larger circle to the edge of the smaller circle is the radius of the larger circle minus the radius of the smaller circle.

### Step 4: Use the Cosine Rule
In triangle \( OAC \), we know that it is equilateral with angles of \( 120^\circ \) between any two sides. Using the cosine rule:
\[ AC^2 = OA^2 + OC^2 - 2 \cdot OA \cdot OC \cdot \cos(120^\circ) \]
Substituting the values:
\[ (2a)^2 = (1 - a)^2 + (1 - a)^2 - 2(1 - a)(1 - a)(-\frac{1}{2}) \]

### Step 5: Simplify the Equation
Expanding the equation:
\[ 4a^2 = 2(1 - a)^2 + (1 - a)^2 \]
\[ 4a^2 = 2(1 - 2a + a^2) + (1 - 2a + a^2) \]
\[ 4a^2 = 3(1 - 2a + a^2) \]
\[ 4a^2 = 3 - 6a + 3a^2 \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ 4a^2 - 3a^2 + 6a - 3 = 0 \]
\[ a^2 + 6a - 3 = 0 \]

### Step 7: Apply the Quadratic Formula
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \]
\[ a = \frac{-6 \pm \sqrt{36 + 12}}{2} \]
\[ a = \frac{-6 \pm \sqrt{48}}{2} \]
\[ a = \frac{-6 \pm 4\sqrt{3}}{2} \]
\[ a = -3 \pm 2\sqrt{3} \]

### Step 8: Select the Positive Solution
Since \( a \) must be positive, we choose:
\[ a = -3 + 2\sqrt{3} \]

### Final Answer
Thus, the value of \( a \) is:
\[ a = 2\sqrt{3} - 3 \]