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**Hint:**All the possible values that x can take are known as the domain, that is the domain of the function is defined as the values of the x for which a function exists. We get different values of y by putting different values of x, thus the range is defined as the set of all the possible values that a function can attain. Using the above-mentioned definition of domain and range of a function, we can find out the domain and range of the given function.

**Complete step-by-step solution:**

We are given that ${x^2} + {y^2} = 9$

We can rewrite it as –

\[y = \sqrt {9 - {x^2}} \]

For y to exist,

$

9 - {x^2} \geqslant 0 \\

\Rightarrow 9 \geqslant {x^2} \\

\Rightarrow {x^2} \leqslant 9 \\

$

Now square rooting both sides and then using the law of inequalities, we get –

$x \geqslant - 3,\,x \leqslant 3$

So, the domain of ${x^2} + {y^2} = 9$ is $[ - 3,3]$

To find the range, we will express x in terms of y –

$x = \sqrt {9 = {y^2}} $

The expression is similar to the one for finding the domain, so the range of ${x^2} + {y^2} = 9$ is $[ - 3,3]$ .

Hence, the domain and the range of ${x^2} + {y^2} = 9$ is $[ - 3,3]$

**Note:**For solving this kind of question, we must know the concept of the domain and range of function clearly. We are given the equation of a circle. The equation of the circle having centre at the origin is given as ${x^2} + {y^2} = {r^2}$ , where r is the radius of the circle, so the radius of the given circle is 3 units. Thus, the maximum values that x and y can take are -3 and 3. So the domain and the range of the function ${x^2} + {y^2} = 9$ is $[ - 3,3]$.

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