Fusing relationships

I've also found an issue within your equations, x^2+y^2=r^2 does not make a circle, it will create an oval. When r=1, x^2+y^2=1. This creates an oval, not a circle...
 
JoshuaSass said:
I've also found an issue within your equations, x^2+y^2=r^2 does not make a circle, it will create an oval. When r=1, x^2+y^2=1. This creates an oval, not a circle...
Click to expand...
If you plot:

x^2 + y^2 = 1 (1^2) on a plane graph paper, it will describe a circle with radius 1.

Can you explain the difference between an oval and a circle?
 
JoshuaSass said:
I've also found an issue within your equations, x^2+y^2=r^2 does not make a circle, it will create an oval. When r=1, x^2+y^2=1. This creates an oval, not a circle...
Click to expand...
Here is the graph of [imath]\left(x^{2}+y^{2}-1\right)\left(y-2x\right)=0[/imath] in Desmos:
1776631960640.png
Looks like a circle to me. @fixit86 has exactly the right idea.

If what you mean by "oval" is "ellipse", then the general form (with center at the origin) is [imath]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/imath]; a circle is a special case, with [imath]a=b=r[/imath]. So a circle is an oval, in this sense. If an "oval" has to have [imath]a\ne b[/imath], then the equation is for a circle, not an oval.
 
Ah, I was imputing it incorrectly, that is my fault. But I'm still confused upon what the OP is asking here?
 
There is no explicit question. He may have been asking for confirmation that you can obtain the union of two graphs ("fusing" them) by multiplying the equations (equated to zero), as in his [imath]\left(x^{2}+y^{2}-1\right)\left(y-2x\right)=0[/imath]. I did so by showing the graph.
 
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