The value of `theta` in `[0, 2pi]` so that circle `x^(2)+y^(2)+2 (sin alpha)x+2(cos alpha)y+sin^(2)theta=0` always lies inside the square of unit side length, is/are

Options

A. `(pi//3, 2pi//3)`
B. `[4pi//3, 5pi//3]`
C. `(pi//4, 2pi//3)`
D. none of these

Answer

To find the values of \( \theta \) in the interval \([0, 2\pi]\) such that the circle given by the equation

\[ x^2 + y^2 + 2(\sin \alpha)x + 2(\cos \alpha)y + \sin^2 \theta = 0 \]

always lies inside a square of unit side length, we can follow these steps:

### Step 1: Rewrite the Circle Equation
The equation of the circle can be rewritten in standard form. The general form of a circle is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

From the given equation, we can identify the center and radius. The equation can be rearranged as:

\[ x^2 + y^2 + 2(\sin \alpha)x + 2(\cos \alpha)y = -\sin^2 \theta \]

Completing the square for \(x\) and \(y\):

\[ (x + \sin \alpha)^2 + (y + \cos \alpha)^2 = \sin^2 \theta + \sin^2 \alpha + \cos^2 \alpha \]

Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\):

\[ (x + \sin \alpha)^2 + (y + \cos \alpha)^2 = 1 + \sin^2 \theta \]

### Step 2: Identify the Center and Radius
From the rewritten equation, we can see that the center of the circle is at \((- \sin \alpha, - \cos \alpha)\) and the radius \(r\) is given by:

\[ r = \sqrt{1 + \sin^2 \theta} \]

### Step 3: Condition for the Circle to Lie Inside the Square
For the circle to lie entirely within the unit square (with vertices at \((0,0)\), \((1,0)\), \((1,1)\), and \((0,1)\)), the maximum distance from the center of the circle to any side of the square must be greater than or equal to the radius of the circle.

The maximum distance from the center \((- \sin \alpha, - \cos \alpha)\) to any side of the square is:

\[ \text{Distance to } x = 0: \quad \sin \alpha \]
\[ \text{Distance to } x = 1: \quad 1 + \sin \alpha \]
\[ \text{Distance to } y = 0: \quad \cos \alpha \]
\[ \text{Distance to } y = 1: \quad 1 + \cos \alpha \]

The minimum of these distances must be greater than or equal to the radius:

\[ \min(\sin \alpha, 1 + \sin \alpha, \cos \alpha, 1 + \cos \alpha) \geq \sqrt{1 + \sin^2 \theta} \]

### Step 4: Finding the Condition for \( \theta \)
The maximum radius of the circle that can fit inside the square is \( \frac{1}{2} \). Thus, we require:

\[ \sqrt{1 + \sin^2 \theta} \leq \frac{1}{2} \]

Squaring both sides gives:

\[ 1 + \sin^2 \theta \leq \frac{1}{4} \]

This leads to:

\[ \sin^2 \theta \leq -\frac{3}{4} \]

Since \(\sin^2 \theta\) is always non-negative, this condition cannot be satisfied. Therefore, we need to find the values of \( \theta \) such that:

\[ |\cos \theta| \leq \frac{1}{2} \]

### Step 5: Solve for \( \theta \)
The values of \( \theta \) that satisfy \( |\cos \theta| \leq \frac{1}{2} \) are:

1. From \( \cos \theta = \frac{1}{2} \):
- \( \theta = \frac{\pi}{3} \)
- \( \theta = \frac{5\pi}{3} \)

2. From \( \cos \theta = -\frac{1}{2} \):
- \( \theta = \frac{2\pi}{3} \)
- \( \theta = \frac{4\pi}{3} \)

Thus, the intervals for \( \theta \) where the circle lies within the square are:

\[ \theta \in \left[\frac{\pi}{3}, \frac{2\pi}{3}\right] \cup \left[\frac{4\pi}{3}, \frac{5\pi}{3}\right] \]

### Final Answer
The values of \( \theta \) in \([0, 2\pi]\) such that the circle always lies inside the square are:

\[ \theta \in \left[\frac{\pi}{3}, \frac{2\pi}{3}\right] \cup \left[\frac{4\pi}{3}, \frac{5\pi}{3}\right] \]