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How do you find the domain and range of ${x^2} + {y^2} = 9$ ?

Last updated date: 21st Feb 2024
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IVSAT 2024
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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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