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.Center: (−2,2), Radius: 1.
Calculate New Center Coordinates: Calculate the new coordinates of the center after rotation by π/6. Rotation matrix R(π/6)=[cos(π/6)−sin(π/6)sin(π/6)cos(π/6)] =[3/2−1/21/23/2]. New center = R(π/6)⋅[−2,2]=[(−2)(3/2)+(2)(−1/2),(−2)(1/2)+(2)(3/2)] =[−3+(−1),−1+3] $= [-\sqrt{\(3\)} - \(1\), \sqrt{\(3\)} - \(1\)].
Write Rotated Circle Equation: Write the equation of the rotated circle using the new center.\(\newline\)Equation: \((x + (\sqrt{3} + 1))^2 + (y - (\sqrt{3} - 1))^2 = 1\).