Check whether the followings represent function or not (i) `x^2 + y^2 = 36, y in [0, 6]`

Explanation

To determine whether the equation \( x^2 + y^2 = 36 \) with the constraint \( y \in [0, 6] \) represents a function, we will follow these steps:

### Step 1: Identify the Equation
The given equation is:
\[ x^2 + y^2 = 36 \]
This represents a circle with a center at the origin (0, 0) and a radius of 6.

### Step 2: Understand the Concept of Function
A relation is a function if for every input \( x \) there is exactly one output \( y \). In other words, for each value of \( x \), there should be only one corresponding value of \( y \).

### Step 3: Rearranging the Equation
We can rearrange the equation to express \( y \) in terms of \( x \):
\[ y^2 = 36 - x^2 \]
Taking the square root gives:
\[ y = \sqrt{36 - x^2} \quad \text{and} \quad y = -\sqrt{36 - x^2} \]

### Step 4: Applying the Constraint
Since we have the constraint \( y \in [0, 6] \), we only consider the non-negative root:
\[ y = \sqrt{36 - x^2} \]
This means for each \( x \) in the domain, \( y \) will take on values from 0 to 6.

### Step 5: Checking Values of \( x \)
To check if every value of \( x \) gives a unique \( y \), we can test specific values of \( x \):
- For \( x = 0 \):
\[ y = \sqrt{36 - 0^2} = \sqrt{36} = 6 \]
- For \( x = 2 \):
\[ y = \sqrt{36 - 2^2} = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2} \]
- For \( x = 6 \):
\[ y = \sqrt{36 - 6^2} = \sqrt{36 - 36} = 0 \]

### Step 6: Conclusion
Since for each value of \( x \) in the domain, there is exactly one corresponding value of \( y \) (within the range of [0, 6]), we conclude that the relation represented by the equation \( x^2 + y^2 = 36 \) with \( y \in [0, 6] \) is indeed a function.

### Final Answer
Yes, the given relation represents a function.

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