Write the rectangular equation in polar form.
(x – 4)2 + y2 = 16
(x-4)^2+y^2=16
x^2-8x+16+y^2=16
x^2+y^2=8x
r^2=8r*cos(theta)
r=8cos(theta)
(x–4)2
49
+
Y2
16
=1?
To write the rectangular equation (x – 4)² + y² = 16 in polar form, we can use the following steps:
Step 1: Replace x and y with their respective polar forms.
Let x = rcosθ and y = rsinθ, where r is the distance from the origin and θ is the angle with the positive x-axis.
Step 2: Substitute the values of x and y into the equation and simplify.
Substituting x = rcosθ and y = rsinθ into (x – 4)² + y² = 16, we get:
(rcosθ – 4)² + (rsinθ)² = 16
Expanding the equation:
r²cos²θ – 8rcosθ + 16 + r²sin²θ = 16
Simplifying:
r²cos²θ + r²sin²θ – 8rcosθ = 0
Using the trigonometric identity cos²θ + sin²θ = 1, we can rewrite the equation as:
r² - 8rcosθ = 0
Step 3: Solve for r to obtain the polar form of the equation.
Factoring out r, we have:
r(r - 8cosθ) = 0
Therefore, the polar form of the equation is:
r = 0 or r = 8cosθ
To convert the given rectangular equation (x – 4)^2 + y^2 = 16 to polar form, we need to express the equation in terms of polar coordinates (r, θ).
In polar form, the Cartesian coordinates (x, y) are represented by (r, θ), where r is the distance from the origin to the point and θ is the angle formed between the positive x-axis and the line connecting the origin to the point.
To convert to polar form, we substitute the Cartesian coordinates with their corresponding polar forms:
(x – 4)^2 + y^2 = 16
Substitute x = r * cos(θ) and y = r * sin(θ):
(r * cos(θ) – 4)^2 + (r * sin(θ))^2 = 16
Expand and simplify:
(r^2 * cos^2(θ) – 8r * cos(θ) + 16) + (r^2 * sin^2(θ)) = 16
Since cos^2(θ) + sin^2(θ) = 1, we can rewrite the equation as:
r^2 * (cos^2(θ) + sin^2(θ)) – 8r * cos(θ) + 16 = 16
Simplify further:
r^2 – 8r * cos(θ) + 16 = 0
This is the polar form of the given equation, expressed in terms of the polar coordinates (r, θ).