Points on circle graph (AMC 10A 2015 #12)

Question

I'm stumped by the following question:

Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^2+x^4=2x^2y+1$. What is $|a-b|$?

Method

See, my train of thought followed as such:

First of all, we see that the two points have the same x value, and thus they probably lie on the circumference of a circle.

$y^{ 2 }+x^{ 4 }=2x^{ 2 }y+1$

$z = { x }^{ 2 }$

$y^{ 2 }-2zy+z^{ 2 }=1$

This simplifies to: ${ (y-z) }^{ 2 }=1$

$y=1+z$

We remember that x = $\sqrt { \pi } $, and since $z={ x }^{ 2 }$, $z=\pi $, y = $1+\pi $. This is where I realized I went wrong somewhere, since |a-b| = 0 at this point. I really have no idea how to proceed at this point, as I believed the problem originally required basic algebraic manipulation.

Answer 1

You made a mistake when you concluded that $(y-z)^2=1 \implies y-z=1$. In fact, $$(y-z)^2=1 \implies y-z=\pm1 \implies y=z\pm 1.$$ Finally, $|a-b|=|z+1-(z-1)|=2$.

Answer 2

we have $$x^4-2x^2y+y^2=1$$ or $$(x^2-y)^2=1$$ plugging the coordinates we get $$\pi-a)^2=1$$ and $$(\pi-b)^2=1$$