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


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.
