## moongazer 5 years ago The equation of a circle center at the origin and radius 1is ?

1. anonymous

x^2 +y^2 =1

2. moongazer

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

3. anonymous

there isnt anyway of explaining it really lol

4. anonymous

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

5. anonymous

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

6. anonymous

i can explain

7. anonymous

ok lets go back to geometry definition

8. anonymous

you there?

9. moongazer

ok

10. anonymous

ohh its phytagorus , yeh

11. moongazer

yes

12. anonymous

the locus definition

13. anonymous

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

14. anonymous

right

15. anonymous

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

16. anonymous

thats the ancient greek definition, going back when

17. anonymous

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

18. anonymous

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

19. anonymous

ok so far?

20. anonymous

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

21. anonymous

ill do it

22. moongazer

ok but please explain easier to understand

23. anonymous

he wants a thorough explanation

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

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

29. moongazer

please explain it like you are explaining it to a kid

30. anonymous

then square both sides

31. anonymous

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

32. anonymous

right that works

33. moongazer

WAIT!

34. 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

35. anonymous

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

36. anonymous

this is a deductive proof my friend :)

37. 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

38. moongazer

on my ntbk

39. anonymous

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

40. anonymous

you get pythagorean triangle

41. moongazer

please don't give me a complicated solution

42. anonymous

ok youre annoying, bye

43. moongazer

then?

44. moongazer

youre doing it right

45. anonymous

then what?

46. anonymous

lols

47. anonymous

then the world ends tomorrow

48. anonymous

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

49. moongazer

ahhhh ok

50. anonymous

a more interesting question has peaked my curiosity, brb

51. moongazer

ill just give you both a medal^_^

52. moongazer

53. anonymous

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

54. anonymous

this question is ambiguous, ok one sec

55. anonymous

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

56. moongazer

what is this?

57. anonymous

58. anonymous

its pretty basic

59. moongazer

it is alright know^_^