What is the image of a circle with equation (x + 2)^2 + (y − 2)^2=1 when it is rotated through π/6 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 original center and radius of the circle. The given equation is (x + 2)^2 + (y − 2)^2 = 1, which means the center is (-2, 2) and the radius is 1. Calculate New Center Coordinates: Calculate the new coordinates of the center after rotation by π/6. Rotation formula: x' = x cos(θ) - y sin(θ), y' = x sin(θ) + y cos(θ). Here, θ = π/6, cos(π/6) = √3/2, sin(π/6) = 1/2. Plugging in 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 using the new center coordinates and the same radius. The new equation is: (x + (√3 + 1))^2 + (y - (√3 - 1))^2 = 1.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

What is the image of a circle with equation $(x + 2)^2 + (y - 2)^2=1$ when it is rotated through $\frac{\pi}{6}$ about the origin? $\newline$ (a) $(x + (\sqrt{3} + 1))^2 + (y - ( \sqrt{3} - 1))^2 = 1$ $\newline$ (b) $(x - ( \sqrt{3} + 1))^2 + (y + (\sqrt{3} - 1))^2 = 1$ $\newline$ (c) $(x + (\sqrt{3} - 1))^2 + (y - ( \sqrt{3} + 1))^2 = 1$ $\newline$ (d) $(x - ( \sqrt{3} - 1))^2 + (y + (\sqrt{3} + 1))^2 = 1$

View step-by-step help

Home

Math Problems

Algebra 2

Reflections of functions

Full solution

Q. What is the image of a circle with equation

(x + 2)^2 + (y - 2)^2=1

when it is rotated through

\frac{\pi}{6}

about the origin?

\newline

(a)

(x + (\sqrt{3} + 1))^2 + (y - ( \sqrt{3} - 1))^2 = 1

\newline

(b)

(x - ( \sqrt{3} + 1))^2 + (y + (\sqrt{3} - 1))^2 = 1

\newline

(c)

(x + (\sqrt{3} - 1))^2 + (y - ( \sqrt{3} + 1))^2 = 1

\newline

(d)

(x - ( \sqrt{3} - 1))^2 + (y + (\sqrt{3} + 1))^2 = 1

Identify Center and Radius: Identify the original center and radius of the circle. The given equation is $(x + 2)^2 + (y - 2)^2 = 1$ , which means the center is $(-2, 2)$ and the radius is $1$ .
Calculate New Center Coordinates: Calculate the new coordinates of the center after rotation by $\pi/6$ . Rotation formula: $x' = x \cos(\theta) - y \sin(\theta)$ , $y' = x \sin(\theta) + y \cos(\theta)$ . Here, $\theta = \pi/6$ , $\cos(\pi/6) = \sqrt{3}/2$ , $\sin(\pi/6) = 1/2$ . Plugging in $x = -2$ , $y = 2$ : $\newline$ $x' = (-2)(\sqrt{3}/2) - (2)(1/2) = -\sqrt{3} - 1$ , $\newline$ $y' = (-2)(1/2) + (2)(\sqrt{3}/2) = -1 + \sqrt{3}$ .
Write Rotated Circle Equation: Write the equation of the rotated circle using the new center coordinates and the same radius. The new equation is: $\newline$ $(x + (\sqrt{3} + 1))^2 + (y - (\sqrt{3} - 1))^2 = 1$ .