P(1, 1) is a point on a circle x2+y2 = 2. If P travels on the circle and reaches Q such that are PQ=112(circumference of the circle), then the coordinates of Q is/are

Given PQ=112(2π2)=π26PQ=rα⇒α=12π26=π6Now ∠XOQ=450±30∘=750or⁡150Q=2Cos⁡750,2Sin⁡750 or 2Cos⁡150,2Sin⁡15∘

P(1, 1) is a point on a circle x2+y2 = 2. If P travels on the circle and reaches Q such that are PQ=112(circumference of the circle), then the coordinates of Q is/are

Given PQ=112(2π2)=π26PQ=rα⇒α=12π26=π6Now ∠XOQ=450±30∘=750or⁡150Q=2Cos⁡750,2Sin⁡750 or 2Cos⁡150,2Sin⁡15∘

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P(1, 1) is a point on a circle $x^{2} + y^{2}$ = 2. If P travels on the circle and reaches Q such that are $PQ = \frac{1}{12}$ (circumference of the circle), then the coordinates of Q is/are

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$(\frac{\sqrt{3} - 1}{2}, \frac{\sqrt{3} + 1}{2})$

$(\frac{3 - \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2})$

$(\frac{\sqrt{3} + 1}{2}, \frac{\sqrt{3} - 1}{2})$

$(\frac{3 + \sqrt{3}}{2}, \frac{3 - \sqrt{3}}{2})$

answer is A, C.

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Detailed Solution

$\begin{array}{l} G i v e n PQ = \frac{1}{12} (2 π \sqrt{2}) = \frac{π \sqrt{2}}{6} \\ P Q = rα \Rightarrow α = \frac{1}{\sqrt{2}} (\frac{π \sqrt{2}}{6}) = \frac{π}{6} \\ Now ∠ XOQ = (45^{0} \pm 30^{\circ}) = 75^{0} or 15^{0} \\ Q = [\sqrt{2} Cos 75^{0}, \sqrt{2} Sin 75^{0}] or [\sqrt{2} Cos 15^{0}, \sqrt{2} Sin 15^{\circ}] \end{array}$