The number of points `(a + 1,a)` where `a in I`, lying inside the region bounded `x^(2) + y^(2) -2x -15 =0` is

Options

A. 2
B. 3
C. 1
D. 6

Answer

To solve the problem of finding the number of points \((a + 1, a)\) where \(a \in \mathbb{I}\) (the set of integers) that lie inside the region bounded by the circle defined by the equation \(x^2 + y^2 - 2x - 15 = 0\), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the Circle's Equation**:
The given equation of the circle is:
\[ x^2 + y^2 - 2x - 15 = 0 \]
We can rewrite it in standard form by completing the square for \(x\):
\[ (x^2 - 2x) + y^2 = 15 \]
Completing the square for \(x\):
\[ (x - 1)^2 - 1 + y^2 = 15 \implies (x - 1)^2 + y^2 = 16 \]
This represents a circle centered at \((1, 0)\) with a radius of \(4\).

2. **Substitute the Points**:
We need to check the points \((a + 1, a)\) for integer values of \(a\). We substitute \(x = a + 1\) and \(y = a\) into the circle's equation:
\[ (a + 1 - 1)^2 + (a)^2 < 16 \]
Simplifying this gives:
\[ a^2 + a^2 < 16 \implies 2a^2 < 16 \implies a^2 < 8 \]

3. **Solve the Inequality**:
From \(a^2 < 8\), we can find the range for \(a\):
\[ -\sqrt{8} < a < \sqrt{8} \]
Since \(\sqrt{8} = 2\sqrt{2} \approx 2.83\), we have:
\[ -2.83 < a < 2.83 \]

4. **Determine Integer Values**:
The integer values of \(a\) that satisfy this inequality are:
\[ a = -2, -1, 0, 1, 2 \]
Thus, there are a total of \(5\) integer values of \(a\).

5. **Conclusion**:
Therefore, the number of points \((a + 1, a)\) where \(a \in \mathbb{I}\) lying inside the region bounded by the circle is:
\[ \boxed{5} \]