If the radius of the sector is 30 and the center angle is 60 degrees, the area of the sector is

If the radius of the sector is 30 and the center angle is 60 degrees, the area of the sector is

The area of the sector is 30x30x3.14x60/360 = 471

Given that the sector radius is 8 and the arc length is 12, what arc is the central angle and the sector area

Center angle α = L / r = 12 / 8 = 1.5 (radians)
Sector area s = LR / 2 = 12 × 8 △ 2 = 48

The arc length of 60 degree center angle is 62.8 meters, and the sector area of this arc length is () square meters
(the process is also important)

Perimeter C = 62.8 * 360 / 60 = 62.8 * 4 = 251.2
C=2*3.14*R,R=C/2/3.14=251.2/2/3.14=40
S=3.14*40*40=5024
S sector = 5024 * 60 / 360 = 5024 / 4 = 1256

Find the equation of the circle passing through three points a (- 1, - 1) B (- 8,0) C (0,6), and point out the radius and center of the circle

The coordinates of points a, B and C are substituted into the formula respectively
（-1-x'）²+（-1-y'）²=R²
（-8-x'）²+（0-y'）²=R²
（0-x'）²+（6-y'）²=R²
The radius of x'y'r can be solved, that is, the coordinates of the center of R circle (x ', y')

Find the equation of the circle passing through three points a (0,0), B (- 6,0), C (0,8)

A:
The circle passes through three points a (0,0), B (- 6,0), C (0,8)
Then the center of the circle passes through the vertical bisector of AB, x = (- 6 + 0) / 2 = - 3
It also passes through the vertical bisector of AC, y = (0 + 8) / 2 = 4
So: the center of the circle is (- 3,4)
Radius r satisfies: R & # 178; = Ao & # 178; = (- 3-0) &# 178; + (4-0) &# 178; = 25
So: radius r = 5
So: the equation of circle is (x + 3) &# 178; + (y-4) &# 178; = 25

Solving the equation of the circle passing through a (6,0) B (5, - 3) C (3,1)

Let the circular equation be: x09x ^ 2 + y ^ 2 + DX + ey + F = 0
Substituting 3 points, you can get the following result: X09
1)\x09a1^2+b1^2+da1+eb1+f=0\x09
2)\x09a2^2+b2^2+da2+eb2+f=0\x09
3)\x09a3^2+b3^2+da3+eb3+f=0\x09
1-2:\x09d(a1-a2)+e(b1-b2)=(-a1^2-b1^2+a2^2+b2^2)=A=-2
1-3:\x09d(a1-a3)+e(b1-b3)=(-a1^2-b1^2+a3^2+b3^2)=B=-26
Note: x09d = (a1-a2) (b1-b3) - (A1-A3) (B1-B2) = - 10
\x09Dd=A(b1-b3)-B(b1-b2)=80
\x09De=B(a1-a2)-A(a1-a3)=-20
Then there are: x09d = DD / D = - 8
\x09e=De/D=2
\x09f=-a1^2-b1^2-da1-eb1=12
So the equation is: x09x ^ 2 + y ^ 2-8x + 2Y + 12 = 0

Find the equations of the following circles: passing through three points a (- 2.4), B (- 1.3), C (2.6)

Let the equation of circle be x ^ 2 + y ^ 2 + A * x + b * y + C = 0
Substitute the coordinates of a, B and C into
4 + 16-2 * a + 4 * B + C = 0
1+9-a+3*b+c=0
4+36+2*a+6*b+c=0
A = 0, B = - 10, C = 20
The equation of circle is
x^2+y^2-10*y+20=0

Through three points a (- 2,4), B (- 1,3), C (2,6), find the equation of circle
How much will LIANLI get in the end?

The circular equation is: (x-a) ^ 2 + (y-b) ^ 2 = C
Simultaneous equations: (a + 2) ^ 2 + (4-b) ^ 2 = C
(a+1)^2+(3-b)^2=c
(2-a)^2+(6-b)^2=c
Very simple solution equation solution: a = 0, B = 5, C = 5, the circular equation is: x ^ 2 + (Y-5) ^ 2 = 5
This is a circle with (0,5) as its center and sqrt (5) as its radius

The standard equation of the circle passing through points a (4,1), B (- 6,3), C (3,0) is______ ．

Let the equation of a circle be x2 + Y2 + DX + ey + F = 0, and 17 + 4D + e + F = 045 − 6D + 3E + F = 09 + 3D + F = 0 can be obtained by crossing points a (4,1), B (- 6,3) and C (3,0). The solution is d = 1, e = - 9, f = - 12x2 + Y2 + x-9y-12 = 0, so the answer is: (x + 12) 2 + (Y − 92) 2 = 652

The equation of the circle whose center is C (3, - 5) and tangent to the line x-7y + 2 = 0 is______ ．

∵ the distance from the center of a circle to the tangent d = R, that is, r = D = | 3 + 35 + 2 | 12 + 72 = 42, the center of a circle C (3, - 5), the equation of circle C is (x-3) 2 + (y + 5) 2 = 32. So the answer is: (x-3) 2 + (y + 5) 2 = 32