Question

What is the general form of the equation for the given circle centered at O(0, 0)? x2 + y2 + 41 = 0 x2 + y2 − 41 = 0 x2 + y2 + x + y − 41 = 0 x2 + y2 + x − y − 41 = 0

Answer 1

I would say the second one.
x^2 + y^2 = 41

Answer 2

Answer:

The  general form of the equation for the given circle centered at O(0, 0) is:

                                $x^2+y^2-41=0$

Step-by-step explanation:

We know that the standard form of circle is given by:

$(x-h)^2+(y-k)^2=r^2$

where the circle is centered at (h,k) and the radius of circle is: r units

1)

$x^2+y^2+41=0$

i.e. we have:

$x^2+y^2=-41$

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)

$x^2+y^2-41=0$

It could also be written as:

$x^2+y^2=41$

which is also represented by:

$(x-0)^2+(y-0)^2=(\sqrt{41})^2$

This means that the circle is centered at (0,0).

3)

$x^2+y^2+x+y-41=0$

It could be written in standard form by:

$(x+\dfrac{1}{2})^2+(y+\dfrac{1}{2})^2=(\sqrt{\dfrac{83}{2}})^2$

Hence, the circle is centered at $(-\dfrac{1}{2},-\dfrac{1}{2})$

Hence, option: 3 is incorrect.

4)

$x^2+y^2+x-y=41$

In standard form it could be written by:

$(x+\dfrac{1}{2})^2+(y-\dfrac{1}{2})^2=(\sqrt{\dfrac{83}{2})^2$

Hence, the circle is centered at:

$(\dfrac{-1}{2},\dfrac{1}{2})$