Question

Find the parametric representation of the circle $3{{x}^{2}}+3{{y}^{2}}+4x-6y-4=0$.

Answer 1

Hint: The parametric equation of a circle is $\begin{aligned} & x=r\cos t+{{x}_{0}} \\\ & y=r\sin t+{{y}_{0}} \\\ \end{aligned}$, where $({{x}_{0}},{{y}_{0}})$ is the centre of the circle, and $r$ is the radius of the circle. However, make the equation given match the general form of the equation of a circle first.

Complete step by step answer:
The general equation of a circle is : ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, where its centre C = $(-g,-f)$ and its radius $r$ =$\sqrt{{{g}^{2}}+{{f}^{2}}-c}$.
The same equation above, can be represented in its parametric form as :
$\begin{aligned} & x=r\cos t+g \\\ & y=r\sin t+f \\\ \end{aligned}$ ………………………..(1)
Where $t$ is the angle the line joining the centre and the point makes with the positive $x$ axis. Usually, if there are no restrictions imposed on the location of the point, $0\le t\le 2\pi$.
Before comparing the given equation to the general form of the circle, it’ll be better if we make the coefficients of ${{x}^{2}}$ and ${{y}^{2}}$ equal to 1, since then it’ll match the general equation’s form exactly, with $g$, $f$ and $c$ being the variables to be determined.
Dividing $S_1$ : $3{{x}^{2}}+3{{y}^{2}}+4x-6y-4=0$ by 3 on both sides, we get :
$S_1:{{x}^{2}}+{{y}^{2}}+\dfrac{4}{3}x-2y-\dfrac{4}{3}=0$
Now, the equation is fit to be compared with the general form. Doing so, we get :

& 2gx=\dfrac{4}{3}x \\\ & \Rightarrow 2g=\dfrac{4}{3} \\\ & \Rightarrow g=\dfrac{2}{3} \\\ \end{aligned}$$ and $$\begin{aligned} & 2fy=-2y \\\ & \Rightarrow 2f=-2 \\\ & \Rightarrow f=-1 \\\ \end{aligned}$$ Hence, the centre $C_1$ of $S_1$ = $(-g,-f)=(-\dfrac{2}{3},1)$ and its radius $r_1$=$\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{\dfrac{{{2}^{2}}}{{{3}^{2}}}+{{1}^{2}}+\dfrac{4}{3}}=\sqrt{\dfrac{4}{9}+1+\dfrac{4}{3}}=\sqrt{\dfrac{4+9+12}{9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}$ Substituting for $g,f,r$ in equation (1), we get : $\begin{aligned} & x=r\cos t+g \\\ & \Rightarrow x=\dfrac{5}{3}\cos t+\dfrac{2}{3} \\\ \end{aligned}$ and $\begin{aligned} & y=r\sin t+f \\\ & \Rightarrow y=\dfrac{5}{3}\sin t-1 \\\ \end{aligned}$ Therefore, our required parametric equation is : $\begin{aligned} & x=\dfrac{5}{3}\cos t+\dfrac{2}{3} \\\ & y=\dfrac{5}{3}\sin t-1 \\\ \end{aligned}$ , $0\le t\le 2\pi $ The parametric equation is : $\begin{aligned} & x=\dfrac{5}{3}\cos t+\dfrac{2}{3} \\\ & y=\dfrac{5}{3}\sin t-1 \\\ \end{aligned}$ , $0\le t\le 2\pi $ Note: Attaching the condition on $t$ is absolutely necessary, since this makes sure that there are no restrictions on the angle that the radius line can make with the positive $x$ axis. Thus, it makes sure that the point can lie anywhere on the circle.