`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 solve the problem, we need to find the coordinates of point Q on the circle defined by the equation \( x^2 + y^2 = 4 \), given that point P is \( (\sqrt{2}, \sqrt{2}) \) and the arc PQ represents \( \frac{1}{4} \) of the circumference of the circle.

### Step-by-Step Solution:

1. **Identify the Circle's Center and Radius:**
The equation of the circle is \( x^2 + y^2 = 4 \).
- The center (O) of the circle is at the origin \( (0, 0) \).
- The radius (r) of the circle is \( \sqrt{4} = 2 \).

**Hint:** The center of the circle can be found from the standard form of the circle's equation.

2. **Determine the Angle Subtended by Arc PQ:**
Since the arc PQ is \( \frac{1}{4} \) of the circumference, the angle \( \angle POQ \) subtended at the center O is:
\[ \angle POQ = 90^\circ \quad (\text{or } \frac{\pi}{2} \text{ radians}) \]

**Hint:** The circumference of a circle is \( 2\pi \), and \( \frac{1}{4} \) of that corresponds to \( 90^\circ \).

3. **Find the Slope of OP:**
The coordinates of point P are \( (\sqrt{2}, \sqrt{2}) \).
- The slope of line OP (from O to P) is:
\[ \text{slope of OP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sqrt{2} - 0}{\sqrt{2} - 0} = 1 \]

**Hint:** The slope formula is used to find the angle of inclination of line segments.

4. **Determine the Slope of OQ:**
Since \( OP \) and \( OQ \) are perpendicular, the slope of OQ will be the negative reciprocal of the slope of OP:
\[ \text{slope of OQ} = -\frac{1}{\text{slope of OP}} = -1 \]

**Hint:** The product of the slopes of two perpendicular lines is -1.

5. **Set Up the Equation for Q:**
Let the coordinates of point Q be \( (h, k) \). Since the slope of OQ is -1, we have:
\[ \frac{k - 0}{h - 0} = -1 \implies k = -h \]

**Hint:** Use the slope definition to relate the coordinates of point Q.

6. **Use the Circle Equation for Q:**
Since Q lies on the circle, we substitute \( k = -h \) into the circle's equation:
\[ h^2 + k^2 = 4 \implies h^2 + (-h)^2 = 4 \implies 2h^2 = 4 \implies h^2 = 2 \implies h = \pm \sqrt{2} \]

**Hint:** Substitute the relationship found in the previous step into the circle's equation.

7. **Find the Corresponding k Values:**
- If \( h = \sqrt{2} \), then \( k = -\sqrt{2} \).
- If \( h = -\sqrt{2} \), then \( k = \sqrt{2} \).

**Hint:** Use the relationship \( k = -h \) to find the corresponding y-coordinates.

8. **Determine the Coordinates of Q:**
The possible coordinates for Q are:
- \( Q_1 = (\sqrt{2}, -\sqrt{2}) \)
- \( Q_2 = (-\sqrt{2}, \sqrt{2}) \)

**Hint:** List all possible combinations based on the values of h and k.

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