The parametric equations `x=(2a(1-t^(2)))/(1+t^(2))` and `y=(4at)/(1+t^(2))` represents a circle whose radius is

Options

A. a
B. 2a
C. 3a
D. 4a

Answer

To find the radius of the circle represented by the parametric equations \( x = \frac{2a(1-t^2)}{1+t^2} \) and \( y = \frac{4at}{1+t^2} \), we will follow these steps:

### Step 1: Substitute \( t \) with \( \tan(\theta) \)
We will use the trigonometric substitution \( t = \tan(\theta) \). This gives us:
- \( t^2 = \tan^2(\theta) \)
- \( 1 - t^2 = 1 - \tan^2(\theta) = \frac{\cos^2(\theta) - \sin^2(\theta)}{\cos^2(\theta)} \)
- \( 1 + t^2 = 1 + \tan^2(\theta) = \frac{1}{\cos^2(\theta)} \)

### Step 2: Rewrite \( x \) and \( y \)
Using the substitution, we can rewrite \( x \) and \( y \):
\[ x = \frac{2a(1 - \tan^2(\theta))}{1 + \tan^2(\theta)} = 2a \cdot \cos(2\theta) \]
\[ y = \frac{4a\tan(\theta)}{1 + \tan^2(\theta)} = 2a \cdot \sin(2\theta) \]

### Step 3: Square and add \( x \) and \( y \)
Now we square \( x \) and \( y \) and add them:
\[ x^2 + y^2 = (2a \cos(2\theta))^2 + (2a \sin(2\theta))^2 \]
\[ = 4a^2 \cos^2(2\theta) + 4a^2 \sin^2(2\theta) \]
Using the Pythagorean identity \( \cos^2(2\theta) + \sin^2(2\theta) = 1 \):
\[ x^2 + y^2 = 4a^2 (\cos^2(2\theta) + \sin^2(2\theta)) = 4a^2 \]

### Step 4: Compare with the standard equation of a circle
The equation \( x^2 + y^2 = 4a^2 \) can be rewritten as:
\[ x^2 + y^2 = (2a)^2 \]
This matches the standard form of a circle \( x^2 + y^2 = r^2 \), where \( r \) is the radius.

### Conclusion
From the comparison, we can see that the radius \( r \) of the circle is:
\[ r = 2a \]

### Final Answer
The radius of the circle is \( 2a \).

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