Geometric interpretation of $ \frac{x^2+y^2}{y}=\text{constant} $.

Question

I would love help in interpreting the following expression geometrically

$$ \frac{x^2+y^2}{y}=\text{constant} $$

for simplicity, let $ c = \text{constant} $, and then through rearrangement we have

$$ x^2+y^2=cy $$

where it is evident that

$$ cy = r^2 $$

where r represents radius

Conceptually, I've broken the problem down into the radius being a function of y, but I am having trouble moving forward from this point. How should I proceed?

Answer 1

The equation is equivalent to:

$$x^2+(y-c/2)^2=c^2/4, \ \ y\neq 0.$$

Does this look familiar? One thing to keep in mind is your original equation does not allow for $y=0$, so this carries over to the above equation, a circle missing points on the $x$ axis.

Answer 2

Note that $$x^2+y^2=cy\iff x^2+(y-c/2)^2=c^2/4.$$ So, the solution is a circumference, without the points where $y=0.$ If $c\le 0$ you can only consider $y\ne 0$, because we can't divide by zero.

Answer 3

Let's say we have the following equation, as you did: $$x^2+y^2=cy$$

Subtract both sides by $cy$: $$x^2+y^2-cy=0$$

We want to complete the square with $y^2-cy$. To do this, add both sides by $\left(\frac c 2\right)^2$. $$x^2+y^2-cy+\left(\frac c 2\right)^2=\left(\frac c 2\right)^2$$

$y^2-cy+\left(\frac c 2\right)^2=\left(y-\frac c 2\right)^2$, so substitute on the left side:

$$x^2+\left(y-\frac c 2\right)^2=\left(\frac c 2\right)^2$$

Thus, this is simply a circle centered at $\left(0, \frac c 2\right)$ with radius $\frac c 2$.