# Is it possible to define this explicitly?

## Main Question or Discussion Point

(x-3)^2 + (y-2)^2 = 1

With some rearranging I get

y(y-4) = -x^2+6x-13

Is it possible to define this explicitly in terms of y? It looks like a mathematical impossibility to define it explicitly in terms of y but I am hoping there might be some analytical technique for this.

## Answers and Replies

pwsnafu
What does "define explicitly" even mean? The formula at the start explicitly defines a circle.
Do you want a function $y(x)$? Then it should be easy to see the answer is no: for every x there are two values of y.

What does "define explicitly" even mean? The formula at the start explicitly defines a circle.
Do you want a function $y(x)$? Then it should be easy to see the answer is no: for every x there are two values of y.
Define the function explicitly in terms of y.

And okay.

AlephZero
Homework Helper
I don't see why not being single-valued means it's impossible to define something explicitly.

$(y-2)^2 = \text{something}$
$(y-2) = \pm \sqrt{\text{something}}$
...

I don't see why not being single-valued means it's impossible to define something explicitly.

$(y-2)^2 = \text{something}$
$(y-2) = \pm \sqrt{\text{something}}$
...
Well in this case it is more like (y^2-4y) = something. so it is bit more complicated. Who knows, maybe one day some genius might develop a clever theory to solve it.

pwsnafu
I don't see why not being single-valued means it's impossible to define something explicitly.

$(y-2)^2 = \text{something}$
$(y-2) = \pm \sqrt{\text{something}}$
...
This is why I asked what "define explicitly" means.

Well in this case it is more like (y^2-4y) = something. so it is bit more complicated. Who knows, maybe one day some genius might develop a clever theory to solve it.
How did you get that?
Edit: And why not just complete the square?

Mark44
Mentor
(x-3)^2 + (y-2)^2 = 1

With some rearranging I get

y(y-4) = -x^2+6x-13

Is it possible to define this explicitly in terms of y?
You can solve the first equation for y in terms of x. However, the equation defines a relation between x and y, but doesn't represent a function. Some values of x map to two different y values.

Having said all that, it's pretty simple to solve for y. Leave the terms in y on one side, and move the terms in x to the other side. Do not expand the (y - 2)2 term. Then take the square root of both sides, remembering to use ± for the two solutions.
It looks like a mathematical impossibility to define it explicitly in terms of y but I am hoping there might be some analytical technique for this.

You can solve the first equation for y in terms of x. However, the equation defines a relation between x and y, but doesn't represent a function. Some values of x map to two different y values.

Having said all that, it's pretty simple to solve for y. Leave the terms in y on one side, and move the terms in x to the other side. Do not expand the (y - 2)2 term. Then take the square root of both sides, remembering to use ± for the two solutions.
Oh, now I see what Alephzero was saying.