The tangentto the curve `x^2 + y^2 - 2x- 3 = 0` is parallel to x-axis at the points

Options

A. `(2,pmsqrt3)`
B. `(1,pm2)`
C. `(pm1,2)`
D. `(pm3,0)`

Answer

To solve the problem of finding the points on the curve \(x^2 + y^2 - 2x - 3 = 0\) where the tangent is parallel to the x-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Curve**:
The given equation is \(x^2 + y^2 - 2x - 3 = 0\). We can rearrange it to identify the type of curve:
\[ x^2 - 2x + y^2 = 3 \]
Completing the square for \(x\):
\[ (x - 1)^2 + y^2 = 4 \]
This represents a circle with center at \((1, 0)\) and radius \(2\).

2. **Determine the Condition for Tangents**:
A tangent to the curve is parallel to the x-axis when its slope is \(0\). The slope of the tangent line can be found using implicit differentiation.

3. **Differentiate the Curve**:
Differentiate the equation \(x^2 + y^2 - 2x - 3 = 0\) with respect to \(x\):
\[ 2x + 2y \frac{dy}{dx} - 2 = 0 \]
Rearranging gives:
\[ 2y \frac{dy}{dx} = 2 - 2x \]
Thus,
\[ \frac{dy}{dx} = \frac{2 - 2x}{2y} = \frac{1 - x}{y} \]

4. **Set the Slope to Zero**:
For the tangent to be parallel to the x-axis:
\[ \frac{1 - x}{y} = 0 \]
This implies \(1 - x = 0\), hence \(x = 1\).

5. **Substitute \(x\) back into the Curve**:
Substitute \(x = 1\) into the original curve equation to find \(y\):
\[ (1)^2 + y^2 - 2(1) - 3 = 0 \]
Simplifying gives:
\[ 1 + y^2 - 2 - 3 = 0 \implies y^2 - 4 = 0 \]
Thus,
\[ y^2 = 4 \implies y = \pm 2 \]

6. **Identify the Points**:
The points where the tangent is parallel to the x-axis are:
\[ (1, 2) \quad \text{and} \quad (1, -2) \]

### Final Answer:
The points on the curve where the tangent is parallel to the x-axis are \((1, 2)\) and \((1, -2)\).

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