0

In finding Cartesian coordinates to $r = 3\sin t$, I found

$$\frac{y^2}{x^2 + y^2} = \frac{x^2 + y^2}{9}.$$

After graphing it with a computer, I see it's a circle. I then managed to rewrite the equation to $x^2 + y^2 - 3y = 0$, but with my level of sophistication, I wasn't even sure this was a circle. But it is! It's a circle of diameter 3 centered what seems to be (0,\sqrt{3}).

But how come? Isn't a circle $(x - a)^2 + (y - b)^2 = r^2$? Where's my $r^2$ in $x^2 + y^2 - 3y = 0$? Please educate me. Thank you!

UPDATE $1$: after some excellent answers, I see the circle is centered around 1.5 and not $\sqrt3$. What I was missing is completing the square (in the $y$ term) so as to write the circle in its standard form. I don't know which excellent answer should be chosen as the answer.
Joep Awinita
  • 145

4 Answers4

2

The terms in $y$ can be written as $$y^2-3y=y^2-3y+(3/2)^2-(3/2)^2=(y-3/2)^2-(3/2)^2$$

This is called 'completing the square', as in realizing that $a^2+2ab+b^2=(a+b)^2$ would work if you interpret $y^2$ to be the $a^2$, and the $-3y$ to be the $2ab$.

Then, completing the square consists in adding (and subtracting) what should be the term $b^2$.

1

Completing the square we have

$$x^2 + y^2 - 3y = x^2+\left(y-\frac32\right)^2-\frac 94=0\implies x^2+\left(y-\frac32\right)^2=\frac 94$$

then

  • center is in $(0,3/2)$
  • radius $R=\frac32$
user
  • 154,566
1

Note that $\left\{(x,y)\ |\ \frac{y^2}{x^2 + y^2} = \frac{x^2 + y^2}{9}\right\}$ is in fact the union of two circles. Indeed, the equation $$(x^2+y^2)^2=9y^2$$ is equivalent to $$x^2+y^2={\color{red}\pm} 3y$$

Which gives you $$x^2+\left(y\ {\color{red} -}\ \frac32\right)^2=\frac 94\qquad \text{and}\qquad x^2+\left(y\ {\color{red} +}\ \frac32\right)^2=\frac 94$$

These circles are symmetric with respect to the $x$-axis, where they are also tangent, thus forming a nice figure eight shape.

However, as noted by J.G., this intermediate equation is not equivalent to the original $r=3\sin \theta$, since $$r=3\sin \theta\Longrightarrow y>0$$ as $y=r\sin \theta=3\sin^2 \theta$.

Arnaud Mortier
  • 27,276
  • Although the original $r=3\sin\theta$ implies $x^2+y^2=r^2=3r\sin\theta=3y$. – J.G. May 30 '18 at 20:34
  • @J.G. True. Thanks, I will edit. – Arnaud Mortier May 30 '18 at 20:38
  • I would say $y > 0$ even without the polar equation. Taking the square root of the cartesian equation we get $|y| = (x^2 + y^2)/3$, so $y$ must be positive since the right hand side is positive. So we can drop the absolute value. Doesn't this proceed? – Joep Awinita May 30 '18 at 20:45
  • @JoepAwinita Definitely no. The RHS being positive tells you that $|y|$ is positive, but doesn't tell you anything about the sign of $y$. That's the meaning of the first part of my post. – Arnaud Mortier May 30 '18 at 20:49
  • @ArnaudMortier You're perfectly right. I conclude the same now. There is no way to tell $y > 0$ unless we look at $r = 3 \sin t$. Thanks for such great contribution. – Joep Awinita May 31 '18 at 14:57
1

$x^2+y^2-3y=0 \implies x^2+(y-\frac{3}{2})^2=(\frac{3}{2})^2$.

So, the equation represents a circle with centre $(0,\frac{3}{2})$ and radius $\frac{3}{2}$.

Surajit
  • 1,216