A circle of diameter 20 units passes through the point `(-3,-1)`. If the centre of a circle of `(2alpha-1,3alpha+1)`, then natural namber `alpha` is

Options

A. 3
B. 4
C. 2
D. 1

Answer

To solve the problem, we need to find the natural number \( \alpha \) such that the center of the circle, given by the coordinates \( (2\alpha - 1, 3\alpha + 1) \), lies on a circle with a diameter of 20 units that passes through the point \((-3, -1)\).

### Step-by-Step Solution:

1. **Find the Radius of the Circle:**
The diameter of the circle is given as 20 units. Therefore, the radius \( r \) can be calculated as:
\[ r = \frac{d}{2} = \frac{20}{2} = 10 \text{ units} \]

**Hint:** Remember that the radius is half of the diameter.

2. **Identify the Center of the Circle:**
Let the center of the circle be \( O \) with coordinates \( (2\alpha - 1, 3\alpha + 1) \) and let the point \( A \) through which the circle passes be \( (-3, -1) \).

3. **Use the Distance Formula:**
The distance \( OA \) (from the center to the point on the circle) should equal the radius. Using the distance formula:
\[ OA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
where \( O = (2\alpha - 1, 3\alpha + 1) \) and \( A = (-3, -1) \).
Thus, we have:
\[ OA = \sqrt{((-3) - (2\alpha - 1))^2 + ((-1) - (3\alpha + 1))^2} \]

4. **Simplify the Expression:**
Substitute the coordinates into the distance formula:
\[ OA = \sqrt{((-3 - 2\alpha + 1)^2 + (-1 - 3\alpha - 1)^2)} \]
This simplifies to:
\[ OA = \sqrt{(-2 - 2\alpha)^2 + (-2 - 3\alpha)^2} \]
\[ = \sqrt{(2(1 + \alpha))^2 + (2(1 + \frac{3}{2}\alpha))^2} \]
\[ = \sqrt{4(1 + \alpha)^2 + 4(1 + \frac{3}{2}\alpha)^2} \]
\[ = 2\sqrt{(1 + \alpha)^2 + (1 + \frac{3}{2}\alpha)^2} \]

5. **Set the Distance Equal to the Radius:**
Since \( OA = 10 \):
\[ 2\sqrt{(1 + \alpha)^2 + (1 + \frac{3}{2}\alpha)^2} = 10 \]
Dividing both sides by 2:
\[ \sqrt{(1 + \alpha)^2 + (1 + \frac{3}{2}\alpha)^2} = 5 \]

6. **Square Both Sides:**
Squaring both sides gives:
\[ (1 + \alpha)^2 + (1 + \frac{3}{2}\alpha)^2 = 25 \]

7. **Expand and Combine Like Terms:**
Expanding both squares:
\[ (1 + 2\alpha + \alpha^2) + (1 + 3\alpha + \frac{9}{4}\alpha^2) = 25 \]
Combine like terms:
\[ 2 + 5\alpha + \left(1 + \frac{9}{4}\right)\alpha^2 = 25 \]
\[ \frac{13}{4}\alpha^2 + 5\alpha - 23 = 0 \]

8. **Multiply by 4 to Eliminate the Fraction:**
\[ 13\alpha^2 + 20\alpha - 92 = 0 \]

9. **Factor the Quadratic Equation:**
Factoring gives:
\[ (13\alpha - 26)(\alpha + 4) = 0 \]

10. **Solve for \( \alpha \):**
Setting each factor to zero:
\[ 13\alpha - 26 = 0 \Rightarrow \alpha = 2 \]
\[ \alpha + 4 = 0 \Rightarrow \alpha = -4 \text{ (not a natural number)} \]

11. **Conclusion:**
The only natural number solution for \( \alpha \) is:
\[ \alpha = 2 \]

### Final Answer:
The natural number \( \alpha \) is \( 2 \).