`P(sqrt2,sqrt2)` is a point on the circle `x^2+y^2=4` and Q is another point on the circle such that arc PQ=`1/4` circumference. The co-ordinates of Q are

Options

A. `(-sqrt2,-sqrt2)`
B. `(sqrt2,-sqrt2)`
C. `(-sqrt2,sqrt2)`
D. none of these

Answer

To find the coordinates of point Q on the circle \( x^2 + y^2 = 4 \) given that point P is \( (\sqrt{2}, \sqrt{2}) \) and arc PQ is \( \frac{1}{4} \) of the circumference of the circle, we can follow these steps:

### Step 1: Determine the radius of the circle
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
From this, we can see that the radius \( r \) of the circle is:
\[ r = \sqrt{4} = 2 \]

### Step 2: Calculate the circumference of the circle
The circumference \( C \) of the circle can be calculated using the formula:
\[ C = 2\pi r \]
Substituting the value of \( r \):
\[ C = 2\pi \times 2 = 4\pi \]

### Step 3: Find the length of arc PQ
Since arc PQ is \( \frac{1}{4} \) of the circumference, we calculate:
\[ \text{Length of arc PQ} = \frac{1}{4} \times 4\pi = \pi \]

### Step 4: Determine the angle corresponding to arc PQ
The angle \( \theta \) corresponding to arc PQ can be found using the relationship between arc length and radius:
\[ \text{Arc Length} = r \times \theta \]
Substituting the known values:
\[ \pi = 2 \times \theta \implies \theta = \frac{\pi}{2} \]

### Step 5: Find the parametric angle for point P
The coordinates of point P are given as \( (\sqrt{2}, \sqrt{2}) \). In parametric form, the coordinates of any point on the circle can be expressed as:
\[ (x, y) = (r \cos \theta, r \sin \theta) \]
For point P:
\[ \sqrt{2} = 2 \cos \theta \quad \text{and} \quad \sqrt{2} = 2 \sin \theta \]
From these equations, we can find:
\[ \cos \theta = \frac{\sqrt{2}}{2} \quad \text{and} \quad \sin \theta = \frac{\sqrt{2}}{2} \]
This corresponds to:
\[ \theta = \frac{\pi}{4} \]

### Step 6: Determine the parametric angle for point Q
Since arc PQ is \( \frac{1}{4} \) of the circumference, the angle for point Q will be:
\[ \text{Angle for Q} = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \]
Alternatively, Q could also be at:
\[ \text{Angle for Q} = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4} \]

### Step 7: Calculate the coordinates of point Q for both angles
1. For \( \theta = \frac{3\pi}{4} \):
\[ Q = (2 \cos \frac{3\pi}{4}, 2 \sin \frac{3\pi}{4}) = \left(2 \cdot -\frac{1}{\sqrt{2}}, 2 \cdot \frac{1}{\sqrt{2}}\right) = (-\sqrt{2}, \sqrt{2}) \]

2. For \( \theta = -\frac{\pi}{4} \):
\[ Q = (2 \cos -\frac{\pi}{4}, 2 \sin -\frac{\pi}{4}) = \left(2 \cdot \frac{1}{\sqrt{2}}, 2 \cdot -\frac{1}{\sqrt{2}}\right) = (\sqrt{2}, -\sqrt{2}) \]

### Final Answer
The coordinates of point Q are:
1. \( Q_1 = (-\sqrt{2}, \sqrt{2}) \)
2. \( Q_2 = (\sqrt{2}, -\sqrt{2}) \)