The equation of the unit circle is given by:
\[ x^2 + y^2 = 1 \]
In this case, we know \( y = 2\sqrt{2} \). To find the value of \( x \), we can substitute \( y \) into the equation of the unit circle:
\[ x^2 + (2\sqrt{2})^2 = 1 \]
Calculating \( (2\sqrt{2})^2 \):
\[ (2\sqrt{2})^2 = 4 \cdot 2 = 8 \]
So we substitute this back into the equation:
\[ x^2 + 8 = 1 \]
Now, we can isolate \( x^2 \):
\[ x^2 = 1 - 8 \]
\[ x^2 = -7 \]
Since \( x^2 = -7 \) is not possible for real numbers, it indicates that the point \( (x, y) \) with \( y = 2\sqrt{2} \) does not intersect the unit circle.
Thus, there is no value of \( x \) for point \( (x, y) \) on the unit circle corresponding to \( y = 2\sqrt{2} \) in Quadrant I.
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