Determine whether the following represents a circle , a point or no circle :
(i) ` 1- x^(2) - y^(2) = 0 ` (ii) `x^(2) + y^(2) + 2x + 1 = 0 ` (iii) `x^(2) + y^(2) - 3x + 3y + 10 = 0 ` .

Answer

To determine whether the given equations represent a circle, a point, or no circle, we will analyze each equation step by step using the general form of a circle's equation.

### General Form of a Circle
The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Where:
- The center of the circle is \((-g, -f)\)
- The radius \(r\) is given by:
\[ r = \sqrt{g^2 + f^2 - c} \]

### Step-by-Step Solution

#### (i) Equation: \(1 - x^2 - y^2 = 0\)

1. **Rearrange the equation:**
\[ x^2 + y^2 = 1 \]
This can be rewritten as:
\[ x^2 + y^2 + 0x + 0y - 1 = 0 \]
Here, \(g = 0\), \(f = 0\), and \(c = -1\).

2. **Calculate the radius:**
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{0^2 + 0^2 - (-1)} = \sqrt{1} = 1 \]
Since \(r > 0\), this represents a circle.

#### (ii) Equation: \(x^2 + y^2 + 2x + 1 = 0\)

1. **Rearrange the equation:**
\[ x^2 + y^2 + 2x + 1 = 0 \]
This can be rewritten as:
\[ x^2 + y^2 + 2x + 0y + 1 = 0 \]
Here, \(g = 1\), \(f = 0\), and \(c = 1\).

2. **Calculate the radius:**
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 0^2 - 1} = \sqrt{0} = 0 \]
Since \(r = 0\), this represents a point.

#### (iii) Equation: \(x^2 + y^2 - 3x + 3y + 10 = 0\)

1. **Rearrange the equation:**
\[ x^2 + y^2 - 3x + 3y + 10 = 0 \]
This can be rewritten as:
\[ x^2 + y^2 - 3x + 3y + 10 = 0 \]
Here, \(g = -\frac{3}{2}\), \(f = \frac{3}{2}\), and \(c = 10\).

2. **Calculate the radius:**
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right)^2 - 10} \]
\[ = \sqrt{\frac{9}{4} + \frac{9}{4} - 10} = \sqrt{\frac{18}{4} - 10} = \sqrt{\frac{18}{4} - \frac{40}{4}} = \sqrt{\frac{-22}{4}} = \sqrt{-\frac{22}{4}} \]
Since \(r < 0\), this represents no circle.

### Final Results
1. The first equation represents a **circle**.
2. The second equation represents a **point**.
3. The third equation represents **no circle**.