The general form of the equation for the given circle centered at O(0, 0) is:
[tex]x^2+y^2-41=0[/tex]
We know that the standard form of circle is given by:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
where the circle is centered at (h,k) and the radius of circle is: r units
1)
[tex]x^2+y^2+41=0[/tex]
i.e. we have:
[tex]x^2+y^2=-41[/tex]
which is not possible.
( Since, the sum of the square of two numbers has to be greater than or equal to 0)
Hence, option: 1 is incorrect.
2)
[tex]x^2+y^2-41=0[/tex]
It could also be written as:
[tex]x^2+y^2=41[/tex]
which is also represented by:
[tex](x-0)^2+(y-0)^2=(\sqrt{41})^2[/tex]
This means that the circle is centered at (0,0).
3)
[tex]x^2+y^2+x+y-41=0[/tex]
It could be written in standard form by:
[tex](x+\dfrac{1}{2})^2+(y+\dfrac{1}{2})^2=(\sqrt{\dfrac{83}{2}})^2[/tex]
Hence, the circle is centered at [tex](-\dfrac{1}{2},-\dfrac{1}{2})[/tex]
Hence, option: 3 is incorrect.
4)
[tex]x^2+y^2+x-y=41[/tex]
In standard form it could be written by:
[tex](x+\dfrac{1}{2})^2+(y-\dfrac{1}{2})^2=(\sqrt{\dfrac{83}{2})^2[/tex]
Hence, the circle is centered at:
[tex](\dfrac{-1}{2},\dfrac{1}{2})[/tex]
