Consider the circle `x^(2) + y^(2) – 2x – 2y – 14=0`, which of following is/are true :

Options

A. Point (3, 4) lies insides the circle
B. Point (–5, –3) lies outside the circle
C. Line x + 3y + 4 = 0 is diameter of circle
D. Area of circle is `16pi` square units

Answer

To solve the given problem regarding the circle defined by the equation \(x^2 + y^2 - 2x - 2y - 14 = 0\), we will first rewrite the equation in standard form and then analyze the options provided.

### Step 1: Rewrite the Circle Equation in Standard Form
The equation of the circle is given as:
\[ x^2 + y^2 - 2x - 2y - 14 = 0 \]
We can rearrange this equation to group the \(x\) and \(y\) terms:
\[ x^2 - 2x + y^2 - 2y = 14 \]
Next, we complete the square for the \(x\) and \(y\) terms.

#### Completing the square for \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

#### Completing the square for \(y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

Substituting these back into the equation gives:
\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 = 14 \]
\[ (x - 1)^2 + (y - 1)^2 - 2 = 14 \]
\[ (x - 1)^2 + (y - 1)^2 = 16 \]

### Step 2: Identify the Center and Radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \((h, k) = (1, 1)\)
- Radius \(r = \sqrt{16} = 4\)

### Step 3: Analyze the Options
Now we will analyze the options provided in the question.

**Option A: Check the point (3, 4)**
We will substitute \(x = 3\) and \(y = 4\) into the equation:
\[ s_1 = 3^2 + 4^2 - 2(3) - 2(4) - 14 \]
Calculating:
\[ s_1 = 9 + 16 - 6 - 8 - 14 = 25 - 28 = -3 \]
Since \(s_1 < 0\), the point (3, 4) lies **inside** the circle.

**Option B: Check the point (-5, -3)**
We will substitute \(x = -5\) and \(y = -3\) into the equation:
\[ s_1 = (-5)^2 + (-3)^2 - 2(-5) - 2(-3) - 14 \]
Calculating:
\[ s_1 = 25 + 9 + 10 + 6 - 14 = 50 - 14 = 36 \]
Since \(s_1 > 0\), the point (-5, -3) lies **outside** the circle.

**Option C: Check if the line \(x + 3y + 4 = 0\) is a diameter**
The center of the circle is (1, 1). We substitute this into the line equation:
\[ 1 + 3(1) + 4 = 1 + 3 + 4 = 8 \neq 0 \]
Since the center does not satisfy the line equation, the line is **not** a diameter of the circle.

**Option D: Check the radius calculation**
The radius was calculated as \(4\) from the standard form. The radius is indeed correct.

### Conclusion
- **Option A**: True (Point (3, 4) is inside the circle)
- **Option B**: True (Point (-5, -3) is outside the circle)
- **Option C**: False (The line is not a diameter)
- **Option D**: True (Radius is correctly calculated)