11x2=-y2+I , I ε Z
i.e x2+y2=I
it is the family of circles with center at origin.
Origin is also a point on the graph (i.e) it can be considered as a circle of 0 radius.
If you understood some of the questions that we have done so far.. then this one will be easy as well...
{x2}={-y2}
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16 Answers
11
62{x2}={-y2}
thus
x2=-y2 + I
thus
x2 + y2 = I
So foro various integers there wll be a circle..
for I=0 there will be a point!
3{x2}={-y2}
which implies
dat x2+y2=I (I is integer)
therefore the graph will be a family of circles having centre at origin and having radius of √I
62to answer honey's question...
What is the solution of {x}=.5
it gives all values x=.5+I
for different integers... (Does this give a hint?)
11No pink for my graph, bhaiyyah? Pls tell me where I went wrong [2]
Radius of each circle is the square root of an integer so that the 'I' in x2+y2=I (I is the square of the radius) is an integer.....
1yup i got it wrong nd realised it ysterday while i ws studying.........so i hd deleted my post.......btw thx
24I think they pink ur post in one thread only once..............I Maybe wrong tooooo..................
1let x2=0.6
and,
then
y2 must be a number with decimal part=0.4
(to satisfy {x2}={-y2})
that means
adding 0.36 to square of any no. with decimal part 0.4 must be
a integer.??
am i correct ??
any1 explain..
62yup the thread is answered :)
The graph abouve is correct mkagenius.. :)