The tangents to the circle `x ^(2) + y ^(2) = 48,` which are inclined at angle of `60^(@)` with the axis of x form a rhombus, the length of whose sides is

Options

A. `2sqrt3`
B. 3
C. `3sqrt3`
D. 4

Answer

To solve the problem of finding the length of the sides of the rhombus formed by the tangents to the circle \(x^2 + y^2 = 48\) that are inclined at an angle of \(60^\circ\) with the x-axis, we can follow these steps:

### Step 1: Determine the radius of the circle
The equation of the circle is given as:
\[ x^2 + y^2 = 48 \]
From this, we can see that the radius \(R\) of the circle is:
\[ R = \sqrt{48} = 4\sqrt{3} \]

### Step 2: Find the slope of the tangents
The tangents to the circle that are inclined at an angle of \(60^\circ\) with the x-axis will have slopes given by:
\[ m = \tan(60^\circ) = \sqrt{3} \]
and
\[ m = \tan(120^\circ) = -\sqrt{3} \]

### Step 3: Write the equations of the tangents
The equation of the tangent line to the circle at a point can be expressed as:
\[ y = mx \pm \sqrt{R^2(1 + m^2)} \]
Substituting \(R = 4\sqrt{3}\) and \(m = \sqrt{3}\):
\[ y = \sqrt{3}x \pm \sqrt{(4\sqrt{3})^2(1 + (\sqrt{3})^2)} \]
Calculating the term under the square root:
\[ (4\sqrt{3})^2 = 48 \quad \text{and} \quad 1 + 3 = 4 \]
Thus:
\[ \sqrt{48 \cdot 4} = \sqrt{192} = 8\sqrt{3} \]
So, the equations of the tangents are:
\[ y = \sqrt{3}x + 8\sqrt{3} \quad \text{and} \quad y = \sqrt{3}x - 8\sqrt{3} \]
And for the negative slope:
\[ y = -\sqrt{3}x + 8\sqrt{3} \quad \text{and} \quad y = -\sqrt{3}x - 8\sqrt{3} \]

### Step 4: Find the intersection points of the tangents
To find the vertices of the rhombus, we need to find the intersection points of the tangents.

1. **Intersection of \(y = \sqrt{3}x + 8\sqrt{3}\) and \(y = -\sqrt{3}x + 8\sqrt{3}\)**:
\[ \sqrt{3}x + 8\sqrt{3} = -\sqrt{3}x + 8\sqrt{3} \]
Simplifying gives:
\[ 2\sqrt{3}x = 0 \implies x = 0 \]
Substituting \(x = 0\) into either equation gives:
\[ y = 8\sqrt{3} \]
So one vertex is \((0, 8\sqrt{3})\).

2. **Intersection of \(y = \sqrt{3}x - 8\sqrt{3}\) and \(y = -\sqrt{3}x - 8\sqrt{3}\)**:
\[ \sqrt{3}x - 8\sqrt{3} = -\sqrt{3}x - 8\sqrt{3} \]
Simplifying gives:
\[ 2\sqrt{3}x = 0 \implies x = 0 \]
Substituting \(x = 0\) into either equation gives:
\[ y = -8\sqrt{3} \]
So the other vertex is \((0, -8\sqrt{3})\).

3. **Intersection of \(y = \sqrt{3}x + 8\sqrt{3}\) and \(y = -\sqrt{3}x - 8\sqrt{3}\)**:
\[ \sqrt{3}x + 8\sqrt{3} = -\sqrt{3}x - 8\sqrt{3} \]
Simplifying gives:
\[ 2\sqrt{3}x = -16\sqrt{3} \implies x = -8 \]
Substituting \(x = -8\) gives:
\[ y = \sqrt{3}(-8) + 8\sqrt{3} = 0 \]
So another vertex is \((-8, 0)\).

4. **Intersection of \(y = \sqrt{3}x - 8\sqrt{3}\) and \(y = -\sqrt{3}x + 8\sqrt{3}\)**:
\[ \sqrt{3}x - 8\sqrt{3} = -\sqrt{3}x + 8\sqrt{3} \]
Simplifying gives:
\[ 2\sqrt{3}x = 16\sqrt{3} \implies x = 8 \]
Substituting \(x = 8\) gives:
\[ y = \sqrt{3}(8) - 8\sqrt{3} = 0 \]
So the last vertex is \((8, 0)\).

### Step 5: Calculate the length of the sides of the rhombus
The distance between the points \((0, 8\sqrt{3})\) and \((0, -8\sqrt{3})\) is:
\[ d = |8\sqrt{3} - (-8\sqrt{3})| = |16\sqrt{3}| \]
The length of a side of the rhombus can be calculated using the distance formula between \((0, 8\sqrt{3})\) and \((-8, 0)\):
\[ \text{Length} = \sqrt{(0 - (-8))^2 + (8\sqrt{3} - 0)^2} = \sqrt{8^2 + (8\sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16 \]

### Final Answer
The length of each side of the rhombus is:
\[ \boxed{16} \]