What is the image of a circle with equation (x+2)2+(y−2)2=1 when it is rotated through 6π about the origin?(a) (x+(3+1))2+(y−(3−1))2=1(b) (x−(3+1))2+(y+(3−1))2=1(c) (x+(3−1))2+(y−(3+1))2=1(d) (x−(3−1))2+(y+(3+1))2=1
Q. What is the image of a circle with equation (x+2)2+(y−2)2=1 when it is rotated through 6π about the origin?(a) (x+(3+1))2+(y−(3−1))2=1(b) (x−(3+1))2+(y+(3−1))2=1(c) (x+(3−1))2+(y−(3+1))2=1(d) (x−(3−1))2+(y+(3+1))2=1
Identify Center and Radius: Identify the center and radius of the original circle. The equation (x+2)2+(y−2)2=1 represents a circle with center (−2,2) and radius 1.
Calculate New Coordinates: Calculate the new coordinates of the center after rotation by π/6. Rotation formula: x′=xcos(θ)−ysin(θ), y′=xsin(θ)+ycos(θ). Here, θ=π/6, cos(π/6)=3/2, sin(π/6)=1/2. Substituting x=−2, y=2: x′=(−2)(3/2)−(2)(1/2)=−3−1, y′=(−2)(1/2)+(2)(3/2)=−1+3.
Write Rotated Circle Equation: Write the equation of the rotated circle. The new center is (−3−1,3−1) and the radius remains 1. The equation becomes:(x+3+1)2+(y−(3−1))2=1.
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