Parameterize an equation

Question

The equation is:

$$x^2 + y^2 - 2x - 3 = 0$$

How do I parameterize that? I haven't encountered equations with both $x^2$, $y^2$ and $x$ before and therefore I am not sure of how to handle the problem.

Should I break out an $x$ to form $y^2+x(x-2)=3$ ?

Answer 1

The equation is $$(x-1)^2+y^2=2^2$$ So it's a circle centered at $(1,0)$ and with radius $2$.

So, $$x=1+2\cos\theta$$ $$y=2\sin\theta$$

Answer 2

Rewrite the equation $(x-1)^2+y^2=2^2$ then it is paramaterised as \begin{eqnarray*} x=1+ 2\cos( \theta) \\ y= 2 \sin ( \theta). \\ \end{eqnarray*}

Answer 3

A rational parametrisation:

The equation describes the circle of radius $2$, centred at $(1,0)$. It passes through the point $(-1,0)$. Let's cut the circle with a variable straight line of slope $t$ passing through this point. An equation of this straight line is $\; y=t(x+1)$ and the abscissæ of its intersections with the circle are the roots of the quadratic equation: $$x^2+t^2(x+1)^2-2x-3=0\iff(1+t^2)x^2+2(t^2-1)x+t^2-3=0.$$ One of these roots is $-1$, so the other root is the opposite of their product $\smash[b]{\dfrac{t^2-3}{1+t^2}}$, whence the parametrisation: \begin{cases}x=\dfrac{3-t^2}{1+t^2},\\y=t(x+1)=\dfrac{4t}{1+t^2}.\end{cases}