The radius of the circle passing through the points of intersection of ellipse `x^2/a^2+y^2/b^2=1` and `x^2-y^2 = 0 ` is

Options

A. `(ab)/sqrt(a^(2) + b^(2))`
B. `(sqrt2 ab)/sqrt(a^(2) + b^(2))`
C. `(a^(2) - b^(2))/sqrt(a^(2) + b^(2))`
D. `(a^(2) + b^(2))/sqrt(a^(2) + b^(2)) `

Answer

To find the radius of the circle passing through the points of intersection of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) and the hyperbola \( x^2 - y^2 = 0 \), we can follow these steps:

### Step 1: Find the Points of Intersection
The equation of the hyperbola can be rewritten as:
\[ x^2 = y^2 \]
This implies:
\[ y = \pm x \]
Now we will substitute \( y = x \) and \( y = -x \) into the ellipse equation.

#### Substituting \( y = x \):
\[ \frac{x^2}{a^2} + \frac{x^2}{b^2} = 1 \]
Combining the terms gives:
\[ x^2 \left( \frac{1}{a^2} + \frac{1}{b^2} \right) = 1 \]
Thus,
\[ x^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{a^2 b^2}{a^2 + b^2} \]
So,
\[ x = \pm \frac{ab}{\sqrt{a^2 + b^2}} \]
And since \( y = x \):
\[ y = \pm \frac{ab}{\sqrt{a^2 + b^2}} \]

#### Substituting \( y = -x \):
\[ \frac{x^2}{a^2} + \frac{(-x)^2}{b^2} = 1 \]
This leads to the same equation as above, yielding the same points of intersection:
\[ \left( \pm \frac{ab}{\sqrt{a^2 + b^2}}, \mp \frac{ab}{\sqrt{a^2 + b^2}} \right) \]

### Step 2: Identify the Points of Intersection
The points of intersection are:
\[ \left( \frac{ab}{\sqrt{a^2 + b^2}}, \frac{ab}{\sqrt{a^2 + b^2}} \right), \left( \frac{ab}{\sqrt{a^2 + b^2}}, -\frac{ab}{\sqrt{a^2 + b^2}} \right), \left( -\frac{ab}{\sqrt{a^2 + b^2}}, \frac{ab}{\sqrt{a^2 + b^2}} \right), \left( -\frac{ab}{\sqrt{a^2 + b^2}}, -\frac{ab}{\sqrt{a^2 + b^2}} \right) \]

### Step 3: Equation of the Circle
The circle that passes through these points can be represented by the equation:
\[ x^2 + y^2 = r^2 \]
To find \( r^2 \), we can use one of the points of intersection, say \( \left( \frac{ab}{\sqrt{a^2 + b^2}}, \frac{ab}{\sqrt{a^2 + b^2}} \right) \):
\[ r^2 = \left( \frac{ab}{\sqrt{a^2 + b^2}} \right)^2 + \left( \frac{ab}{\sqrt{a^2 + b^2}} \right)^2 \]
This simplifies to:
\[ r^2 = 2 \left( \frac{a^2 b^2}{a^2 + b^2} \right) = \frac{2a^2 b^2}{a^2 + b^2} \]

### Step 4: Calculate the Radius
Taking the square root gives:
\[ r = \sqrt{\frac{2a^2 b^2}{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}} \sqrt{2} \]

### Final Answer
Thus, the radius of the circle is:
\[ r = \frac{ab \sqrt{2}}{\sqrt{a^2 + b^2}} \]