The equation of a circle center at the origin and radius 1is ?

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The equation of a circle center at the origin and radius 1is ?

Mathematics
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x^2 +y^2 =1
explain why please ill give you a medal if you explained it well
there isnt anyway of explaining it really lol

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Other answers:

the general form a circle with centre (a,b) and radius r is \[(x-a)^2 +(y-b)^2 =r^2\]
here centre is (0,0) yeh.. theres not alot of explaining, you have to remember the formula , thats all you can do
i can explain
ok lets go back to geometry definition
you there?
ok
ohh its phytagorus , yeh
yes
the locus definition
a circle is a set of points such that, the distance from the center is fixed
right
a circle is a locus of points whose distance from a point is constant, that distance is the radius
thats the ancient greek definition, going back when
now fast forward to cartesian world, we have x y graph
your center lets say for convenience is origin (0,0)
ok so far?
your doing it the really long way, not very good explaination lol
ill do it
ok but please explain easier to understand
he wants a thorough explanation
we want d( (x,y) (0,0) = r
the distance between (0,0) and the points (x,y) to be the constant radius
ok so far? , and we know the distance between any two points using the distance formula
elec, this is not easy to understand? i can show a graph , but this interface sucks balls
you can just go straight to the general case centre (a,b) \[\sqrt{(x-a)^2 +(y-b)^2} = r \]
please explain it like you are explaining it to a kid
then square both sides
d ( (x1,y1) (x2,y2) ) = sqrt ( (x1-x2)^2 + (y1-y2)^2)
right that works
WAIT!
so the distance between any arbitrary center (h,k) and the points (x,y) is some distance r. d ( h,k)(x,y) = r
, so sqrt ( (x-h)^2 + ( y - k)^2) = r
this is a deductive proof my friend :)
my teacher gave me a really easy solution for this and i just forgot it and then i wrote the equation x^2+y^2=r^2 and i forgot what it means
on my ntbk
if you draw a line from your center to a point on the circle, and then draw a triangle
you get pythagorean triangle
please don't give me a complicated solution
ok youre annoying, bye
then?
youre doing it right
then what?
lols
then the world ends tomorrow
brb, i will use a whiteboard , this is no good
ahhhh ok
a more interesting question has peaked my curiosity, brb
ill just give you both a medal^_^
for answering and explaining^_^
complex numbers: find a, b such that (-bi)^2 = a^2 ?
this question is ambiguous, ok one sec
here click on this http://www.twiddla.com/542364
what is this?
circle centre (2,3), radius 2
its pretty basic
it is alright know^_^

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