In a polar coordinate system, the center of the circle is at the root number two, and the equation f

Teach you a powerful way to depolarize coordinates:

In Cartesian coordinate systems, we know that the translation of an image will cause a corresponding change in the image equation, for example, the equation becomes y=(x-1) when the image of y=x is translated to the right

In polar coordinate systems, there is a similar phenomenon, but instead of translation, it rotates.

In a polar coordinate system, the equation for a circle with a radius of r and the center of the circle at (r,0) is easily obtained: r=2rcos

And is the circle with the center of the circle at (2, ) passing through the origin point obtained by the rotation of the circle with the center of the circle at (2,0) and the radius of 2?

In Cartesian coordinate systems, translating in an increasing direction to x will change x to x-1, translating in the direction of decreasing x will change x to x+1, and similarly, in polar coordinates, the counterclockwise direction is the increasing direction, so when rotating counterclockwise, it becomes - and vice versa.

So the equation for a circle with the center of the circle at ( 2, ) and the pole is r=2 2cos( -

Remember, this method can only be used for shapes that rotate around the poles, with the mantra "add and subtract".

It is recommended to use the "Geometry Sketchpad", which is very powerful.

In a polar coordinate system, the polar equation for a circle with a known center of ( 0, ) and a radius r is .

2-2ρ0cos(θ-0^2-r^2=0

Substitute the center of the circle (root number 2, pie) into the above formula.

Within the polar coordinates, the equation for the circle with the center of the circle at the pole is p=rsina

Now you can translate the coordinates, where r = root number 2

The result is p = root number 2 * sina - root number 2

Square of the radius of the circle = 2 + pi 2

The equation for a circle in Cartesian coordinate system is, x-2 (1 2)) 2 + y-pi) 2 = 2 + pi 2, and the equation for a circle in a polar coordinate system is, rcos(t) -2 (1 2)] 2 + rsin(t) -pi] 2 = 2 + pi 2

The radius of the circle r = root number (2 + pi 2).

The angle a satisfies tana = pi and the root number 2 circle equation can be expressed as a function of r with respect to the radiator angle c.

r=2rcos(c-a)

The polar coordinate equation for a circular dispersed rock with a circle center of (2, ) and a passing pole is =4cos( - is =-4cos

Therefore, the answer to the case of cherry blossoms is =-4cos

In the polar coordinate system, the polar coordinate equation of a circle with a known center of ( 0, ) and a radius of r is .

2-2ρ0cos(θ-0^2-r^2=0

Substituting the center of the circle (2, ) into the above equation is sufficient.

Cartesian coordinates c,(2cos 3,2sin 3),c(1, 3), the circular equation is: (x-1) 2+(y- 3) 2=5, x= cos, y= sin, cos -1) 2+( sin - 3) 2=5, the circular polar coordinate equation is: 2-4sin( + 6)* =1.

Center of the circle = ( 2, ).

That is, the radius is 2

Convert to Cartesian coordinates: center of the circle = (-2,0).

Therefore, the equation is: (x+ 2) 2+(y) 2=2, i.e., x 2+2 2x+2+y 2=2

Continue to the polar equation:

2+2√2pcosθ=0

i.e.: +2 2cos = 0

Two-coordinate conversion: x= *cos, y= *sin, x 2+y 2= 2

If you don't understand, please ask.

There is a mistake upstairs.,Please look at the picture.。。。

First of all, it is Cartesian coordinates, y=rsina x=rcosa, so: center of the circle(x0,y0) y0=2sin=0 x0=2cospai=-2

The center of the circle (-2,0) is slowed over the original code. Semi-late mode diameter = 2x+2) 2+y 2=2 2=4

rcosa+2)^2+r^2sin^2a=4r^2(cos^2a+sin^2a)+4rcosa+4=4r^2+4rcosa=0

r(r+4cosa)=0

r=-4cosa is the polar coordinate equation.

<> Question Analysis: Point <>

Cartesian coordinates are <>, and straight lines <>

Transform into <>, making <>

<>, the center of the circle is <> "The equation of the circle is <>

Comments: Polar coordinates <>

<> with Cartesian coordinates

The reciprocal relationship is <>, and this problem is first transformed into an equation in the Cartesian coordinate system according to the reciprocal formula, so as to determine the equation of the lower circle, and finally normalized to the polar coordinates.