Reflection of line (x - 1)^2 + y^2 = 4 in y = x

[Monkeyseat]

Find the equation of the graph after the following has been reflected in the line y = x:

(x - 1)^2 + y^2 = 4

Okay, this is kind of like finding the inverse function. I had been re-arranging them to get x on one side, and then swapping x and y over.

Here is what I did:

(x - 1)^2 + y^2 = 4
y^2 = 4 - (x - 1)^2
y^2 = 4 - (x^2 - 2x + 1)
y^2 = 4 - x^2 + 2x - 1
y^2 = 3 - x^2 + 2x
x^2 - 2x = 3 - y^2
x(x-2) = 3 - y^2
x = 3 - y^2/(x - 2)
y = 3 - x^2/(y - 2)

Or is it this?

(x - 1)^2 + y^2 = 4
y^2 = 4 - (x - 1)^2
y = 2 - (x - 1)
x = 3 - y
y = 3 - x

Neither look right to me. Any suggestions?

Thank you.

 

[Deleted member 4993]

Find the equation of the graph after the following has been reflected in the line y = x:

(x - 1)^2 + y^2 = 4

Okay, this is kind of like finding the inverse function. I had been re-arranging them to get x on one side, and then swapping x and y over.

Here is what I did:

(x - 1)^2 + y^2 = 4
y^2 = 4 - (x - 1)^2
y^2 = 4 - (x^2 - 2x + 1)
y^2 = 4 - x^2 + 2x - 1
y^2 = 3 - x^2 + 2x
x^2 - 2x = 3 - y^2
x(x-2) = 3 - y^2
x = 3 - y^2/(x - 2)
y = 3 - x^2/(y - 2)

Or is it this?

(x - 1)^2 + y^2 = 4
y^2 = 4 - (x - 1)^2
y = 2 - (x - 1)<<<< How is that??!!!
________________________________________-
\(\displaystyle y^2 = 4 - (x - 1)^2\)

\(\displaystyle y \, =\, \pm\, \sqrt{ 4 - (x - 1)^2}\)

_______________________________________
x = 3 - y
y = 3 - x <<< This is equation of a straight line. You started with an equation of a circle. It's reflection - cannot become straight-line.

Neither look right to me. Any suggestions?

Thank you.

You really need to understand and be proficient at fundamentals of algebra - before you are ready to solve these types of problems.

 

[Monkeyseat]

\(\displaystyle y^2 = 4 - (x - 1)^2\)

\(\displaystyle y \, =\, \pm\, \sqrt{ 4 - (x - 1)^2}\)

Thanks for the reply. Right, so I need to get x on one side - would I not just be back at where I started as I would have to square each side?

So then I'd be back to:

y^2 = 4 - (x - 1)^2
y^2 = 4 - (x^2 - 2x + 1)
y^2 = 4 - x^2 + 2x - 1
y^2 = 3 - x^2 + 2x
x^2 - 2x = 3 - y^2
x(x-2) = 3 - y^2
x = 3 - y^2/(x - 2)
y = 3 - x^2/(y - 2)

 

[Deleted member 4993]

\(\displaystyle y^2 = 4 - (x - 1)^2\)

\(\displaystyle y \, =\, \pm\, \sqrt{ 4 - (x - 1)^2}\)

Thanks for the reply. Right, so I need to get x on one side - would I not just be back at where I started as I would have to square each side?

So then I'd be back to:

y^2 = 4 - (x - 1)^2
y^2 = 4 - (x^2 - 2x + 1)
y^2 = 4 - x^2 + 2x - 1
y^2 = 3 - x^2 + 2x
x^2 - 2x = 3 - y^2
x(x-2) = 3 - y^2
x = 3 - y^2/(x - 2)
y = 3 - x^2/(y - 2)

The original equation was not in the form of y = f(x).

Why do you think the reflected equation ought to be in that form?

The equation for the reflected curve is simply:

(y - 1)^2 + x^2 = 4................. That's it!!!

The original curve was a circle of radius = 2 and center at (1,0)

The reflected curve is a circle of radius = 2 and center at (0,1)

Draw these two curves along with the line y = x

and see the relevance of this process.

 

[Monkeyseat]

Right thanks - I see it now after drawing the graph.

It's just I have been doing all these kind of a in a 'set method' (I know you shouldn't really) so I was thrown by this one, as silly as it may sound.

So the original line is:

(x - 1)^2 + y^2 = 4

... and its reflected form is:

(y - 1)^2 + x^2 = 4

I have been used to doing all these by re-arranging for x, so I wasn't sure why not to do that this time. Is it because this is the equation of a circle and not in the form y = f(x) so you just swap x and y over?

Thanks.