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

- moongazer

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

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- schrodinger

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- anonymous

x^2 +y^2 =1

- moongazer

explain why please
ill give you a medal if you explained it well

- anonymous

there isnt anyway of explaining it really lol

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## More answers

- anonymous

the general form a circle with centre (a,b) and radius r is \[(x-a)^2 +(y-b)^2 =r^2\]

- anonymous

here centre is (0,0)
yeh.. theres not alot of explaining, you have to remember the formula , thats all you can do

- anonymous

i can explain

- anonymous

ok lets go back to geometry definition

- anonymous

you there?

- moongazer

ok

- anonymous

ohh its phytagorus , yeh

- moongazer

yes

- anonymous

the locus definition

- anonymous

a circle is a set of points such that, the distance from the center is fixed

- anonymous

right

- anonymous

a circle is a locus of points whose distance from a point is constant, that distance is the radius

- anonymous

thats the ancient greek definition, going back when

- anonymous

now fast forward to cartesian world, we have x y graph

- anonymous

your center lets say for convenience is origin (0,0)

- anonymous

ok so far?

- anonymous

your doing it the really long way, not very good explaination lol

- anonymous

ill do it

- moongazer

ok
but please explain easier to understand

- anonymous

he wants a thorough explanation

- anonymous

we want d( (x,y) (0,0) = r

- anonymous

the distance between (0,0) and the points (x,y) to be the constant radius

- anonymous

ok so far? , and we know the distance between any two points using the distance formula

- anonymous

elec, this is not easy to understand? i can show a graph , but this interface sucks balls

- anonymous

you can just go straight to the general case centre (a,b)
\[\sqrt{(x-a)^2 +(y-b)^2} = r \]

- moongazer

please explain it like you are explaining it to a kid

- anonymous

then square both sides

- anonymous

d ( (x1,y1) (x2,y2) ) = sqrt ( (x1-x2)^2 + (y1-y2)^2)

- anonymous

right that works

- moongazer

WAIT!

- anonymous

so the distance between any arbitrary center (h,k) and the points (x,y) is some distance r.
d ( h,k)(x,y) = r

- anonymous

, so sqrt ( (x-h)^2 + ( y - k)^2) = r

- anonymous

this is a deductive proof my friend :)

- moongazer

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

- moongazer

on my ntbk

- anonymous

if you draw a line from your center to a point on the circle, and then draw a triangle

- anonymous

you get pythagorean triangle

- moongazer

please don't give me a complicated solution

- anonymous

ok youre annoying, bye

- moongazer

then?

- moongazer

youre doing it right

- anonymous

then what?

- anonymous

lols

- anonymous

then the world ends tomorrow

- anonymous

brb, i will use a whiteboard , this is no good

- moongazer

ahhhh ok

- anonymous

a more interesting question has peaked my curiosity, brb

- moongazer

ill just give you both a medal^_^

- moongazer

for answering and explaining^_^

- anonymous

complex numbers: find a, b such that (-bi)^2 = a^2 ?

- anonymous

this question is ambiguous, ok one sec

- anonymous

here click on this http://www.twiddla.com/542364

- moongazer

what is this?

- anonymous

circle centre (2,3), radius 2

- anonymous

its pretty basic

- moongazer

it is alright know^_^

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