How can the polar equation P = 2 be transformed into a rectangular equation? I know it's a circle, but I need standard transformation steps

How can the polar equation P = 2 be transformed into a rectangular equation? I know it's a circle, but I need standard transformation steps

p=√(x^2+y^2)
√(x^2+y^2)=2
x^2+y^2=4

Question 1. Does the linear equation PCOS (θ - π / 3) = a have a straight line x = a turning clockwise or counterclockwise?
How to transform the polar equation PCOS (θ - π / 3) = a into a rectangular equation?
Question 3. There are two ways to express the same point, such as (1,3 ^ (1 / 2)) in polar coordinates. But is there any necessary relationship between the positive and negative value of ρ and θ? And is the positive and negative value of ρ related to its quadrant or something

1
The parametric equation PCOS (θ - π / 3) = ax = a, PCOS 0 = PCOS (π / 3 - π / 3) is obtained by turning π / 3 anticlockwise
two
PCOS (θ - π / 3) = a rotate π / 3 clockwise so that θ '= θ - π / 3, PCOS (θ - π / 3) = PCOS, θ' = a, x = a
three
(1, √ 3) in quadrant I, the value of ρ is independent of quadrant. Generally, the distance from point to o (0,0) is taken as ρ, that is, ρ = √ (x ^ 2 + y ^ 2) > 0
Generally, the value of θ is centered on O (0,0), and the clockwise rotation is the positive direction

High school mathematics: polar coordinates
The circular equation with point (1,1) in polar coordinate system as the center and 1 as the radius is?
Write the method and answer, thank you!

Polar coordinates (1,1) are transformed into plane rectangular coordinates (cos1, sin1)
Circular equation: (x-cos1) ^ 2 + (y-sin1) ^ 2 = 1
x^2+y^2-2xcos1-2ysin1=0
Then it is reduced to polar equation: R ^ 2-2rcosacos1-2rsinasin1 = 0
r^2-2rcos(a-1)=0
r=2cos(a-1)

1、 Can polar angle be negative in polar coordinates?
2、 How to express the polar coordinates of the poles? How to take the polar angle when the polar diameter is zero? Is it OK to write (0,0)?

It can be negative. When the polar diameter is zero, the angle may be arbitrary

It is known that the center of the ellipse is O, the major axis and minor axis are 2a and 2b respectively (a > b > 0), a and B are two points on the ellipse, and OA is vertical to ob
(1) 1 / OA ^ 2 + 1 / ob ^ 2 is the fixed value
(2) Find the maximum and minimum of △ AOB area

The area enclosed by the curve | x | + | y | = root 2 is ()

S = (2 radical 2) ^ 2 / 2 = 4

On the derivation of definite integral,

F '(x) = [0, x] ∫ {&; [(x-2t) f (T)] / &; X} DT + (x-2x) f (x) (DX / DX) = [0, x] ∫ f (T) DT XF (x) > 0, so f (x) is an increasing function

The derivative of indefinite integral, the master of advanced mathematics,
What is the result of this derivation? Please give the specific steps and explain which chapter of high number is involved,

The derivation of variable upper limit definite integral is also the content of high school, just use the formula

The mass is evenly distributed in the area surrounded by the cardioid r = a (1-cos θ) (a > 0)
Find the centroid of the area surrounded by the cardioid r = a (1-cos θ) (a > 0). Note: it is the centroid of the area, not the centroid of the cardioid. Use double integral

Let the density be ρ
Then the mass m = ρ∫∫ DS = ρ∫∫ rdrd θ = ρ∫ (0 - > 2 π) d θ∫ (0 - > A (1-cos θ)) RDR = 3 π a ^ 2 ρ / 2
Since the centroid is symmetric about the X axis, the ordinate Y0 of the centroid is 0
x0=ρ∫∫xds /m=ρ∫∫r^2cosθdrdθ /m=ρ∫(0->2π)cosθdθ ∫(0->a(1-cosθ)) r^2dr /m
=(-5πa^3ρ/4) / (3πa^2ρ/2)
=-5a/6
So the core is (- 5A / 6,0)

1 / (2 + sin ^ 2x) DX under the root sign of COS

In the first question, the marking method is ambiguous; in the second question, the substitution method is used, with the root x = t, and then DX = 2tdt, and the original formula is changed to cost * 2tdt for partial integration