Question

The centre and radius of a circle \[ x = 4a \left( \frac{1 - t^2}{1 + t^2} \right), \quad y = \frac{8at}{1 + t^2} \] are respectively

• A. \( (0, 0) \) and \( 3a \) units
• B. \( (0, 0) \) and \( 4a \) units
• C. \( (0, 0) \) and \( 2a \) units
• D. \( (0, 0) \) and \( a \) units

Hint:

For parametric equations of a circle, the center is the point where the parametric equations yield the origin, and the radius is determined by the coefficients of the terms involving \( t \).

Answer 1

The Correct Option is B

Step 1: Recognize the parametric equations.
The given equations are in the parametric form of a circle. These represent a circle, with \( t \) being the parameter.
Step 2: Find the radius and center of the circle.
For the parametric equations of the form \[ x = 4a \left( \frac{1 - t^2}{1 + t^2} \right), \quad y = \frac{8at}{1 + t^2} \] the center is \( (0, 0) \) and the radius is given by \[ \text{Radius} = \frac{4a}{1 + t^2} \quad \text{(from the parametric equations)} \] Since \( t \) can vary, we get the radius as \( 4a \).
Step 3: Conclusion.
Thus, the center is \( (0, 0) \) and the radius is \( 4a \) units.