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myininaya
True/False: A graph that contains just a single point (h,k) can be written as (x-h)^2+(y-k)^2=0 which means it can be seen as a circle with radius 0. Please state what you think. Don't look it up on a website because I can do that if that is what I wanted. :)
Its true...i didnt look it up
That is what I say to but people actually do take the other side on this one. If there is any reason why you think it is true, can you say why?
that is the formula for it , look in your book
the equality only hold for (x,y) = (h,k) for any set of point other than (h, k) this relation is not valid in Real plane.
@experimentX its the formula for it man
formula is just expression of logic.
Some people don't like the idea of calling something with zero size a circle (or any other plane figure), but I don't have a problem with it. If you want to be more precise, it's the limit of a circle as its radius approaches zero.
That is the standard form of an equation for a circle with r=0, so why not? I would be more general and say it's an ellipse of size zero, but I'm a dork like that.
I think it depends on how you define a circle. I would say the radius could be greater than equal to 0. Someone told me you can actually prove that when you have (x-h)^2+(y-k)^2=0 this is a circle.
I left out the word "or"
Depends on if you want a synthetic geometry definition, analytic geometry definition, calculus definition. As far as I'm concerned, It is a circle. It's a circle with r=0.
So, in more plain language (or using a geometry definition), it would be better to call it a point rather than a circle, but as long as you are clear in your definitions and can show your logic is consistent in either case, either way works.
if you let r=0 then you will destroy properties that all 'normal' circles have
You put quotation marks around normal because you do see it as a circle @zarkon , but not a "normal" circle ?
I would call it a degenerate circle
I like "degenerate."
It's a shorter way of saying "circle with radius zero."
perhaps that way we could avoid confusing the difference between these two. (x-h)^2+(y-k)^2=0 a(x-h)^2+b(y-k)^2=0
but isn't a circle just defined as a center with all points on the circumference equadistant fro the center. Hmmmmm center and circumference are the same here.......
I don't see why the center and circumference cannot coincide. There is nothing in the definition of a circle that forbids that.
that depends on what definition you use
I like my circles to have interiors..and the break the plane into two regions (not including the circle itself)
that way all my theorems hold and make sense
Maybe thinking of the equation as a "circle" is what makes it seem confusing. Think of it as a distance function:\[d: \mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}\]\[((x_1,y_1),(x_2,y_2))\longmapsto \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]
Then in a sense, when you have:\[(x-h)^2+(y-k)^2=0\]you are saying "I want the set of all points (x,y) such that the distance between (x,y) and (h,k) is zero."
@joemath314159 the equation of a circle, the distance formula, and the pythagorean theorem are all the same thing.
draw a tangent to a circle with radius zero
@Zarkon why does a tangent need to be defined in order for it to be a circle?
all other circles have that property..and many other properties that are destroyed by having r=0...maybe it should have its own name...like "point"
And even if there is no single unique tangent to that point, a tangent can still be drawn.
Those are properties of some circles, but are they included in the definition of "circle?"
Argument from personal preference isn't valid, I'd think.
most are theorems that require that a circle have an interior
Then that's just too bad. You use those theorems when they apply. If they don't apply, you don't use the theorems. Whether or not particular theorems apply does not change the definition of "circle."