Find the
- Cartesian equation of the circle whose parametric equations are $x = \dfrac{1}{4} \cos \theta$, $y = \dfrac{1}{4} \sin \theta$ and $0 < \theta \leq 2 \pi$.
- parametric equation of the circle $4x^2 + 4y^2 = 9$.
- Given: $\;\;\;$ $x = \dfrac{1}{4} \cos \theta$ $\implies$ $\cos \theta = 4x$ $\;\;\; \cdots \; (1)$
and $y = \dfrac{1}{4} \sin \theta$ $\implies$ $\sin \theta = 4y$ $\;\;\; \cdots \; (2)$
Squaring and adding equations $(1)$ and $(2)$ gives
$\cos^2 \theta + \sin^2 \theta = 16 x^2 + 16 y^2$
i.e. $16x^2 + 16 y^2 = 1$ $\;\;\; \cdots \; (3)$
Equation $(3)$ is the required equation of the circle in Cartesian form. - Given equation of circle is: $\;\;\;$ $4x^2 + 4y^2 = 9$
i.e. $x^2 + y^2 = \dfrac{9}{4}$
i.e. $x^2 + y^2 = \left(\dfrac{3}{2}\right)^2$
Comparing with the standard equation of circle $\;$ $x^2 + y^2 = a^2$ $\;$ gives the radius $= a = \dfrac{3}{2}$
Parametric equations of the circle $\;$ $x^2 + y^2 = a^2$ $\;$ are $\;$ $x = a \cos \theta$; $y = a \sin \theta$
$\therefore$ $\;$ The required parametric equations of the given circle are
$x = \dfrac{3}{2} \cos \theta$; $y = \dfrac{3}{2} \sin \theta$