Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$

Question

Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$

$$c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$$

$$c{x^2+c(y+1)^2}={x^2+y^2-1}$$

$$c{x^2+cy^2+2cy+c}={x^2+y^2-1}\text{ [expanded]}$$

$$1+c=x^2-cx^2+y^2-cy^2-2cy\text{ [moved to other side]}$$

$$1+c=(1-c)x^2+\color{red}{(1-c)y^2-2cy}\text{ [factor, will complete square of red]}$$

$$1+c=(1-c)x^2+\color{red}{(1-c)(y^2-\dfrac{2cy}{c-1}+(\dfrac{2c}{c-1})^2)-(\dfrac{2c}{c-1})^2}\text{ [completed square]}$$

$$1+c+\color{red}{(\dfrac{2c}{c-1})^2}=(1-c)x^2+\color{red}{(1-c)(y-\dfrac{2c}{c-1})^2}\text{ [done]}$$

So we see that this is an ellipse, stretched in the $y$ direction by a factor of $1-c$, and translated both in $x$ and $y$.

$c\neq 1$.

Consider $c=2$

$19=-x^2-(y+16)^2$

$-19=x^2+(y+16)^2$

But this not a circle nor an ellipse. It's supposed to be a circle? What is the issue here?

Answer 1

Here

$1+c=(1-c)x^2+\color{red}{(1-c)(y^2-\dfrac{2cy}{c-1}+(\dfrac{2c}{c-1})^2)-(\dfrac{2c}{c-1})^2}\text{ [completed square]}$

It's not equal to the precedent line..... sign problem and you didnt add the correct square....

By the way put $c=2$ in the first equation, you get :

$x^2+(y+2)^2=1$ which is a circle.

Answer 2

$$ -(1-c)y^2-2cy = -(1-c)\left[y^2+\frac{2c}{1-c}y\right]=-(1-c)\left[\left(y+\frac{c}{1-c}\right)^2-\left(\frac{c}{1-c}\right)^2\right] $$ Not what you have above.

Answer 3

You do not need all those calculations!

Starting from $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$

you get

$\left(x^2+y^2-1\right)-c \left(x^2+(y+1)^2\right)=0$

expand

$-c x^2-c y^2-2 c y-c+x^2+y^2-1=0$

collect

$(1-c) x^2+(1-c) y^2-2 c y-c-1=0$

and divide each term by $c-1$ provided that $c\ne 1$

$x^2+y^2-\dfrac{2 c }{1-c}\,y-\dfrac{1+c}{1-c}=0$

which is the equation of a circle with centre $\left(0,\dfrac{2 c }{1-c}\right)$

and radius $r=\dfrac{1}{|c-1|}$