The number of points on the circle `2(x^(2)+y^(2))=3x` which are at a distance 2 from the point (-2, 1), is

Options

A. 2
B. 0
C. 1
D. none of these

Answer

To solve the problem, we need to find the number of points on the circle defined by the equation \( 2(x^2 + y^2) = 3x \) that are at a distance of 2 from the point (-2, 1).

### Step-by-step Solution:

1. **Rewrite the Circle Equation**:
Start with the given equation of the circle:
\[ 2(x^2 + y^2) = 3x \]
Dividing the entire equation by 2 gives:
\[ x^2 + y^2 - \frac{3}{2}x = 0 \]

2. **Complete the Square**:
To rewrite the equation in standard form, we complete the square for the \(x\) terms:
\[ x^2 - \frac{3}{2}x + y^2 = 0 \]
Completing the square for \(x\):
\[ \left(x - \frac{3}{4}\right)^2 - \frac{9}{16} + y^2 = 0 \]
Rearranging gives:
\[ \left(x - \frac{3}{4}\right)^2 + y^2 = \frac{9}{16} \]
This represents a circle with center \(\left(\frac{3}{4}, 0\right)\) and radius \(r = \frac{3}{4}\).

3. **Identify the Given Point**:
The point we are considering is \(P(-2, 1)\).

4. **Calculate the Distance from Point P to the Center of the Circle**:
The distance \(PC\) from point \(P(-2, 1)\) to the center of the circle \(C\left(\frac{3}{4}, 0\right)\) is calculated using the distance formula:
\[ PC = \sqrt{\left(-2 - \frac{3}{4}\right)^2 + (1 - 0)^2} \]
Simplifying this:
\[ PC = \sqrt{\left(-\frac{8}{4} - \frac{3}{4}\right)^2 + 1^2} = \sqrt{\left(-\frac{11}{4}\right)^2 + 1} = \sqrt{\frac{121}{16} + 1} = \sqrt{\frac{121}{16} + \frac{16}{16}} = \sqrt{\frac{137}{16}} = \frac{\sqrt{137}}{4} \]

5. **Determine the Condition for Distance**:
We need to find points on the circle that are at a distance of 2 from point \(P\). Thus, we need to check if the distance \(PC\) minus the radius of the circle is greater than or equal to 2:
\[ PC - r = \frac{\sqrt{137}}{4} - \frac{3}{4} \]
To find out if there are points at distance 2, we need:
\[ \frac{\sqrt{137}}{4} - \frac{3}{4} \geq 2 \]
This simplifies to:
\[ \sqrt{137} - 3 \geq 8 \implies \sqrt{137} \geq 11 \]
Since \(\sqrt{121} = 11\), we conclude that \(\sqrt{137} > 11\).

6. **Conclusion**:
Since the distance from point \(P\) to the center of the circle is greater than the radius plus 2, point \(P\) lies outside the circle, and thus there are no points on the circle that are at a distance of 2 from point \(P\).

Therefore, the final answer is:
\[ \text{The number of points on the circle at a distance 2 from } (-2, 1) \text{ is } 0. \]