Answer
This curve describes a circle of radius equal to 1, and centered at $(1, 0)$.
Work Step by Step
$r = 2cos(\theta)$
- Knowing that: $rcos(\theta) = x$ and $r^2 = x^2 + y^2$:
Multiply boths sides by $r$:
$r^2 = 2rcos(\theta)$
$x^2 + y^2 = 2x$
$x^2 - 2x + y^2 = 0$
Add 1 to both sides of the equation:
$x^2 -2x + 1 + y^2 = 1$
$(x -1)^2 + y^2 = 1$
This Cartesian equation follows the pattern for a circle:
$(x - x_0)^2 + (y - y_0)^2 = radius^2$
Where: $(x_0, y_0)$ is the position of its center.
Therefore, this curve describes a circle of radius equal to 1, and centered at $(1, 0)$.
