Math  /  Geometry

QuestionProb. 5 Consider these three diagrams. (a) The first diagram depicts a point (p1,q1)\left(p_{1}, q_{1}\right) lying on a circle of radius 1 centered at the point (0,0)(0,0)- which is to say, the unit circle. What are the coordinates of this point? (p1,q1)=(\left(p_{1}, q_{1}\right)=( \qquad \qquad (b) The second diagram depicts a point (p2,q2)\left(p_{2}, q_{2}\right) lying on a circle of radius 2 centered at the point (0,0)(0,0). (Effectively, we've made the previous circle twice as big.) What are the coordinates of this point? (p2,q2)=(\left(p_{2}, q_{2}\right)=( \qquad \qquad (c) The third diagram depicts a point (p3,q3)\left(p_{3}, q_{3}\right) lying on a circle of radius 2 centered at the point (2,1)(-2,1). (Effectively, we've shifted the previous circle 2 units left and 1 unit up.) What are the coordinates of this point? (p3,q3)=(,)\left(p_{3}, q_{3}\right)=(\square, \square)

Studdy Solution
To find (p3,q3)(p_3, q_3), use the transformation for a circle of radius 2 centered at (2,1)(-2,1):
\[ p_3 = -2 + 2 \cdot \cos\left(\frac{2\pi}{3}\right) $ \[ q_3 = 1 + 2 \cdot \sin\left(\frac{2\pi}{3}\right) $
Calculate:
\[ p_3 = -2 + 2 \cdot \left(-\frac{1}{2}\right) = -3 $ \[ q_3 = 1 + 2 \cdot \frac{\sqrt{3}}{2} = 1 + \sqrt{3} $
Thus, (p3,q3)=(3,1+3)(p_3, q_3) = (-3, 1 + \sqrt{3}).
The coordinates for each point are: - (p1,q1)=(12,32)(p_1, q_1) = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) - (p2,q2)=(1,3)(p_2, q_2) = (-1, \sqrt{3}) - (p3,q3)=(3,1+3)(p_3, q_3) = (-3, 1 + \sqrt{3})

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