If the curves `(x ^(2))/(4) + y ^(2) = 1 and (x ^(2))/(a ^(2)) + y ^(2) =1,` for suitable value of 'a', cut on four concyclic points, the equation of the circle passing through these four points is

Options

A. `x ^(2) + y ^(2) =2`
B. `x ^(2) +y ^(2) =1`
C. `x ^(2) +y ^(2) =4`
D. none of these

Answer

To find the equation of the circle passing through the four concyclic points formed by the intersection of the curves \(\frac{x^2}{4} + y^2 = 1\) and \(\frac{x^2}{a^2} + y^2 = 1\), we will follow these steps:

### Step 1: Identify the curves
The first curve is an ellipse given by:
\[ \frac{x^2}{4} + y^2 = 1 \]
This represents an ellipse with semi-major axis \(2\) along the x-axis and semi-minor axis \(1\) along the y-axis.

The second curve is also an ellipse given by:
\[ \frac{x^2}{a^2} + y^2 = 1 \]
This represents an ellipse with semi-major axis \(a\) along the x-axis and semi-minor axis \(1\) along the y-axis.

### Step 2: Set up the equations for intersection
To find the intersection points of these two ellipses, we can set the equations equal to each other. We can express one variable in terms of the other. For example, we can express \(y^2\) from the first equation:
\[ y^2 = 1 - \frac{x^2}{4} \]
Substituting this into the second equation gives:
\[ \frac{x^2}{a^2} + \left(1 - \frac{x^2}{4}\right) = 1 \]
This simplifies to:
\[ \frac{x^2}{a^2} - \frac{x^2}{4} = 0 \]

### Step 3: Factor out \(x^2\)
Factoring out \(x^2\) from the equation:
\[ x^2\left(\frac{1}{a^2} - \frac{1}{4}\right) = 0 \]
This gives us two cases: \(x^2 = 0\) or \(\frac{1}{a^2} - \frac{1}{4} = 0\).

### Step 4: Solve for \(a\)
From \(\frac{1}{a^2} - \frac{1}{4} = 0\), we can solve for \(a\):
\[ \frac{1}{a^2} = \frac{1}{4} \implies a^2 = 4 \implies a = 2 \]
Thus, both ellipses are identical when \(a = 2\).

### Step 5: Find the intersection points
Substituting \(a = 2\) back into either of the original equations, we can find the intersection points. Using:
\[ \frac{x^2}{4} + y^2 = 1 \]
We can find the points of intersection:
1. When \(x = 0\), \(y^2 = 1\) gives points \((0, 1)\) and \((0, -1)\).
2. When \(x = 2\), \(y^2 = 0\) gives point \((2, 0)\).
3. When \(x = -2\), \(y^2 = 0\) gives point \((-2, 0)\).

Thus, the intersection points are \((0, 1)\), \((0, -1)\), \((2, 0)\), and \((-2, 0)\).

### Step 6: Find the equation of the circle
The four points \((0, 1)\), \((0, -1)\), \((2, 0)\), and \((-2, 0)\) lie on a circle. The center of this circle is at the origin \((0, 0)\) and the radius is the distance from the center to any of the points, which is \(1\).

Thus, the equation of the circle is:
\[ x^2 + y^2 = 1 \]

### Summary
The equation of the circle passing through the four concyclic points is:
\[ \boxed{x^2 + y^2 = 1} \]