Determine the equation of the ellipse `x^(2)/a^(2)+y^(2)/b^(2) = 1 ` such that it has the least area but contains the circle `(x-1)^(2)+ y^(2) = 1`.

Answer

To determine the equation of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) that has the least area while containing the circle \( (x-1)^2 + y^2 = 1 \), we can follow these steps:

### Step 1: Understand the Circle
The given circle has its center at \( (1, 0) \) and a radius of \( 1 \). This means the circle extends from \( x = 0 \) to \( x = 2 \) and from \( y = -1 \) to \( y = 1 \).

### Step 2: Determine the Requirements for the Ellipse
For the ellipse to contain the circle, it must be wide enough in the x-direction and tall enough in the y-direction. The ellipse must cover the entire area of the circle.

### Step 3: Set up the Area of the Ellipse
The area \( A \) of the ellipse is given by the formula:
\[ A = \pi a b \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Step 4: Use the Circle's Dimensions
Since the circle has a radius of \( 1 \) and is centered at \( (1, 0) \), the ellipse must satisfy:
- The x-coordinates must range from \( 0 \) to \( 2 \), so \( a \) must be at least \( 1 \) (to cover the distance from the center to the edge of the circle).
- The y-coordinates must range from \( -1 \) to \( 1 \), so \( b \) must be at least \( 1 \).

### Step 5: Express \( b \) in Terms of \( a \)
To minimize the area while ensuring the ellipse contains the circle, we can express \( b \) in terms of \( a \). The ellipse must be centered at the origin, so we can set:
\[ b = \sqrt{1 - e^2} a \]
where \( e \) is the eccentricity of the ellipse.

### Step 6: Set Up the Discriminant Condition
For the ellipse to have the least area, we need to ensure that the discriminant of the quadratic formed by substituting the circle's equation into the ellipse's equation is zero:
\[ b^2 - a^2 = 0 \]
This leads to the relationship:
\[ b^2 = a^2 - a^2 e^2 \]

### Step 7: Substitute and Minimize Area
Substituting \( b^2 = \frac{1}{e^2} \) into the area formula gives:
\[ A = \pi a \left(\frac{1}{e}\right) = \frac{\pi a}{\sqrt{1 - e^2}} \]
To minimize this area, we can differentiate with respect to \( e \) and set the derivative to zero.

### Step 8: Solve for \( e \)
After differentiating and simplifying, we find:
\[ e^2 = \frac{2}{3} \]
Thus, we can find \( a \) and \( b \) using:
\[ b = \frac{1}{\sqrt{e^2}} = \sqrt{\frac{3}{2}}, \quad a = \frac{1}{\sqrt{1 - e^2}} = \sqrt{3} \]

### Step 9: Write the Equation of the Ellipse
Finally, substituting \( a \) and \( b \) back into the ellipse equation gives:
\[ \frac{x^2}{3} + \frac{y^2}{\frac{3}{2}} = 1 \]
Multiplying through by \( 6 \) to eliminate the fractions results in:
\[ 2x^2 + 4y^2 = 6 \]
or simplifying further:
\[ x^2 + 2y^2 = 3 \]

### Final Answer
The equation of the ellipse that contains the circle and has the least area is:
\[ \frac{x^2}{3} + \frac{y^2}{\frac{3}{2}} = 1 \]