If a circle is inscribed on a sector with a central angle of π / 3 and a radius of R, the ratio of the area of the circle to the sector is_____ .

If a circle is inscribed on a sector with a central angle of π / 3 and a radius of R, the ratio of the area of the circle to the sector is_____ .

Let the radius of the circle be r. after drawing, connect the center of the circle with the vertex of the sector. You can see that one right side of the right triangle is r, so its hypotenuse is 2R
2r+r=R,r=1/3R
So the ratio of the area of the circle to the sector is: π * R ^ 2:1 / 6, π R ^ 2 = 2 / 3

How do you know the equation of a circle and a straight line to find the equation of a symmetrical circle about this straight line?
Circular equation (x + 2) & sup2; + (y-6) & sup2; = 1 linear equation 3x-4y + 5 = 0

The key is to find out the symmetric point of the known circle center about the straight line
The symmetrical point is on the vertical line passing through the center of the circle and connecting with the known straight line
Let the equation of vertical straight line equation 3x-4y + 5 = 0 which has been (- 2,6) be:
The intersection of y-6 = - 4 / 3 (x + 2) and linear equation 3x-4y + 5 = 0 is (1,2)
Let (- 2,6) be symmetric with respect to the linear equation 3x-4y + 5 = 0
（m,n）
1=（-2+m)/2,2=(6+n)/2
The solution is: M = 4, n = - 2
The equation of symmetric circle (x-4) & sup2; + (y + 2) & sup2; = 1

How does the general equation of a circle develop into a standard equation of a circle?
X2 + y2-6x + 4Y + 15 = 0, the formula is (x-3) 2 + (y + 2) 2 = - 2
How does x2 + y2-6x + 4Y + 15 = 0 change into (x-3) 2 + (y + 2) 2 = - 2?

It is conditional that a quadratic equation is a circle. First, the coefficients of quadratic terms are equal, that is, the coefficients of x ^ 2 and Y ^ 2 terms are equal,
Secondly, the condition that x ^ 2 + y ^ 2 + ax + by + C = 0 can be expanded into a circle is that C13 does not satisfy the condition
So it can't expand into a circle

Through points a (3,5) and B (- 3,7), and the center of the circle is on the x-axis, the equation of the circle satisfying the condition is obtained

The center of the circle is on the X axis
Let C be (a, 0)
The circular equation is (x-a) ^ 2 + y ^ 2 = R ^ 2
And because the circle passes through points a (3,5) and B (- 3,7)
So (3-A) ^ 2 + 25 = R ^ 2 ------ (1)
(a+3)^2+49=r^2-------(2)
(1) (2) get a = - 2
r^2=50
The circular equation is (x + 2) ^ 2 + y ^ 2 = 50

Change the equation of the following circle into a standard equation, and write the coordinates and radius of the center of the circle
1、x²+y²-6x=0
2、x²+y²-4y-5=0
3、x²+y²-4x-6y+12=0
4、2x²+2y²-4x+8y+5=0

1. (x-3) &# 178; + Y & # 178; = 3 & # 178; Center: (3,0) radius: 32. X & # 178; + (Y-2) &# 178; = 3 & # 178; Center: (0,2) radius: 33. X & # 178; + (Y-2) &# 178; = 3 & # 178; Center: (2,3) radius: 14.2 (x-1) &# 178; + 2 (y + 2) &# 178; = 5 Center: (1, - 2) half

Write the equation for the ellipse under the following conditions
(1) The focus is on the x-axis, the focal length is equal to 4, and passes through the point P (3, - 2, 6)
(2) The focal coordinates are (0, - 4), (0,4), a = 5
(3)a+c=10,a-c=4.
Sorry, guys. I've missed a few words. It's the standard equation for ellipse. I'm sorry

1.x^2/36+y^2/32=1
2.x^2/9+y^2/25=1
3. X ^ 2 / 49 + y ^ 2 / 40 = 1 or x ^ 2 / 40 + y ^ 2 / 49 = 1

The equation of a circle with a point (1,1) in polar coordinate system as its center and a radius of 1 is ()
A. ρ＝2cos(θ−π4)B. ρ＝2sin(θ−π4)C. ρ=2cos（θ-1）D. ρ=2sin（θ-1）

Let the polar coordinates of any point on the circle be (ρ, θ), then from the radius of 1, (ρ cos θ − COSL) 2 + (ρ sin θ − sinl) 2 = 1, the equation is ρ = 2cos (θ - 1)

In polar coordinates, the equation of a circle with (A2, π 2) as its center and A2 as its radius is______ ．

As shown in the figure, ∵ apo is the circumference angle of the diameter Ao of ⊙ o, ∵ apo = π 2. ∵ ρ = ACOS (π 2 − θ) = asin θ. ∵ ρ = asin θ. So the answer is: ρ = asin θ

Solving the equation of circle with point a (4,4 / π) as center and 2 as radius in polar coordinate system
In polar coordinate system!!!!!!!!!

Do you want a polar equation, with cosine theorem
r^2+12-8rcos(θ-π/4)=0

In polar coordinate system, the equation of a circle with radius a of center coordinates (a, P) (a > 0) is?

The parameter equation of circle in polar coordinates is as follows:
x=acosc
y=asinc
C is the angle between the x-axis and the line where the coordinates and the center of the circle are located, and a is the radius. Therefore, the problem can be solved as follows: first, find out the tangent of the angle, that is, Tanc = P / A, and then find out sinc and COSC according to the trigonometric transformation