Question

How do you decide whether the relation ${{x}^{2}}+{{y}^{2}}=1$ defines a function?

Answer 1

For checking whether the given relation defines a function or not, we have to use the basic definition of a function. A function is a binary relation between two sets which relates every element of the first set to exactly one element of the other set. The element of the first set is called the preimage, while the element of the second set is called the image. For checking this condition on the given relation, we have to separate $y$ and express it in the form of $x$.

Complete step by step answer:
The relation given to us in the question is
${{x}^{2}}+{{y}^{2}}=1$
Subtracting $1$ from both the sides, we get
$\begin{aligned} & \Rightarrow {{x}^{2}}+{{y}^{2}}-1=1-1 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-1=0 \\\ & \Rightarrow {{y}^{2}}+{{x}^{2}}-1=0 \\\ \end{aligned}$
Taking $-1$ common from the last two terms, we have
$\Rightarrow {{y}^{2}}-\left( 1-{{x}^{2}} \right)=0$
Now, writing $\left( 1-{{x}^{2}} \right)$ as ${{\sqrt{\left( 1-{{x}^{2}} \right)}}^{2}}$ in the above equation, we get
$\Rightarrow {{y}^{2}}-{{\sqrt{\left( 1-{{x}^{2}} \right)}}^{2}}=0$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. From the above equation, we have $a=y$ and $b=\sqrt{1-{{x}^{2}}}$. So we can write the above equation as
$\Rightarrow \left( y+\sqrt{\left( 1-{{x}^{2}} \right)} \right)\left( y-\sqrt{\left( 1-{{x}^{2}} \right)} \right)=0$
On solving the above equation, we get
$y=-\sqrt{\left( 1-{{x}^{2}} \right)}...........(i)$, and
$y=\sqrt{\left( 1-{{x}^{2}} \right)}...........(ii)$
From the equations (i) and (ii), we can say that for a single value of $x$, we are getting two different values for $y$, one positive and the other negative.
For example, if we put $x=0$ in the equation (i) then the value of $y$ which we get is
$\begin{aligned} & \Rightarrow y=-\sqrt{\left( 1-{{0}^{2}} \right)} \\\ & \Rightarrow y=-1 \\\ \end{aligned}$
Now, on putting $x=0$ in the equation (ii), the value of $y$ is
$\begin{aligned} & \Rightarrow y=\sqrt{\left( 1-{{0}^{2}} \right)} \\\ & \Rightarrow y=1 \\\ \end{aligned}$
So for the value $x=0$, the given relation gives us two distinct values of $y$, which are $1$ and $-1$.
We know that a function is a relation which has a single image corresponding to a pre-image.
But the given relation has two images for each pre-image.

Hence, the given relation ${{x}^{2}}+{{y}^{2}}=1$ does not define a function.

Note: While expressing $y$ in the form of $x$, always prefer to form a standard quadratic equation in $y$ and solve it to obtain its two solutions. Do not directly take the square root, after separating $x$ and $y$. This is because we might forget the negative solution in this case, and conclude that the given relation defines a function.