
The equation $x^{2}+y^{2}=0$ denotes
A. A point
B. A circle
C. $x$-axis
D. $y$-axis
Answer
162k+ views
Hint:We are asked to find what the given equation represents. For this, we find the loci of the given equation. From that we are able to find what the equation represents. In the sections that follow, we'll use this formula to determine a circle's equation or center.
Complete Step by step Solution:
$x^{2}+y^{2}=0$
$\Rightarrow x^{2}=-y^{2}$
The locus is the collection of all points that satisfy the requirements and create geometrical shapes like lines, line segments, circles, curves, etc. As a result, we may state that they can be thought of as locations where the point can be found or moved rather than as a collection of points.
Except at $(0,0)$, a perfect square can never be equal to the negative of another perfect square.
Therefore, the origin, or point $(0,0),$ is the locus of the given equation.
So the correct answer is option(A)
Note:We can also use the following method:-
We know the general form of circle is
$(x-h)^{2}+(y-k)^{2}=r^{2}$
Now we compare the above equation with the given equation $x^{2}+y^{2}=0$, we get
$h=0$, $k=0$ and $r=0$
That means centre = $(0,0)$ and $r=0$
This represents that the centre is the origin and the radius of the circle is zero.
Hence it denotes a point.
Complete Step by step Solution:
$x^{2}+y^{2}=0$
$\Rightarrow x^{2}=-y^{2}$
The locus is the collection of all points that satisfy the requirements and create geometrical shapes like lines, line segments, circles, curves, etc. As a result, we may state that they can be thought of as locations where the point can be found or moved rather than as a collection of points.
Except at $(0,0)$, a perfect square can never be equal to the negative of another perfect square.
Therefore, the origin, or point $(0,0),$ is the locus of the given equation.
So the correct answer is option(A)
Note:We can also use the following method:-
We know the general form of circle is
$(x-h)^{2}+(y-k)^{2}=r^{2}$
Now we compare the above equation with the given equation $x^{2}+y^{2}=0$, we get
$h=0$, $k=0$ and $r=0$
That means centre = $(0,0)$ and $r=0$
This represents that the centre is the origin and the radius of the circle is zero.
Hence it denotes a point.
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