The length of an arc is 31.4, the radius of the circle is 30, then the central angle of the arc is () degrees

The length of an arc is 31.4, the radius of the circle is 30, then the central angle of the arc is () degrees

The circumference of the circle is 2 π * r = 188.4
31.4 / 188.4 = center angle / 360
Get the center angle of the circle = 60 degrees

If the diameter of a circle is 6 cm, then the arc length of 90 degree center angle is () cm

3.14×6×90°/360°
=71 cm

(1) Given the ellipse x ^ 2 + 4Y ^ 2 = 16, find the linear equation of the chord passing through point P (1,1) and bisected by P
(2) Given that the straight line passing through point P (1,1) intersects ellipse x ^ 2 + 4Y ^ 2 = 16 at two points a and B, find the trajectory equation of midpoint ab

1、
Let Y-1 = K (x-1)
y=kx+(1-k)
Substituting
(1+4k²)x²+8k(1-k)x+4(1-k)²-16=0
The abscissa of the midpoint is (x1 + x2) / 2 = - 4K (1-k) / (1 + 4K & sup2;) = 1
-4k+4k²=1+4k²
k=-1/4
So it's x + 4y-5 = 0
2、
Let Y-1 = K (x-1)
y=kx+(1-k)
Substituting
(1+4k²)x²+8k(1-k)x+4(1-k)²-16=0
x=(x1+x2)/2=-4k(1-k)/(1+4k²)
y=kx+(1-k)
y=(y1+y2)/2=k(x1+x2)/2+(1-k)
=(1-k)/(1+4k²)
So x / y = - 4K
Because Y-1 = K (x-1)
So x / y = - 4 (Y-1) / (x-1)
x²-x=-4y²+4y
x²+4y²-x-4y=0

The known circle C: x2 + y2-4x-6y + 9 = 0（
The known circle C: x2 + y2-4x-6y + 9 = 0
(1) If the point Q (x, y) is on the circle C, find the maximum and minimum of X + y;
(2) It is known that the line L passing through point P (3,2) intersects circle C at two points a and B. if P is the midpoint of line AB, the equation of line L is obtained

First draw a graph, let z = x + y, then y = - x + Z. from the graph, we can see that when the line Z is tangent to the circle, Z is the maximum or minimum, then we can get the coordinates of the minimum value and the maximum value. Bring in the line Z, then we can get the Z value

Let C: x square + y square + 4x-6y = 0
If the circle C is symmetric with respect to the line L: a (x-2y) - (2-A) (2x + 3y-4) = 0, find the value of A

Circle (x + 2) & sup2; + (Y-3) & sup2; = 13
Center (- 2,3)
If a circle is symmetrical with respect to a line, the line is the line in which the diameter lies
So over the center of the circle
Substituting x = - 2, y = 3
-8a-(2-a)=0
-8a-2+a=0
a=-2/7

(find positive solution) a straight line passing through the origin intersects the circle x ^ 2 + y ^ 2-4x = 0 at two points a and B. find the trajectory equation of the midpoint m of the chord ab
To process (detailed ah) had better also have some words

The circle equation is (X-2) ^ 2 + y ^ 2 = 2 ^ 2, the center of the circle is (2,0), the radius is 2, so the origin is on the circle, so it is a point
Then for B (x, y), the midpoint of the line between B (x, y) and a is m (x / 2, Y / 2)
Let's transform the circle equation, divide both sides by 4, there are: (x / 2 - 1) ^ 2 + (Y / 2) ^ 2 = 1
So the trajectory equation of m point is (x-1) ^ 2 + y ^ 2 = 1

If we make a chord of circle x ^ 2 + y ^ 2-12x = 0 through the origin, then the trajectory equation of the midpoint of the chord

x^2+y^2-12x=0
Standard form: (X-6) ^ 2 + y ^ 2 = 36
Know the center of a circle a (6,0) through the center of a circle a (6,0) as the vertical line of the string
From the vertical diameter theorem, the perpendicular foot is the midpoint P of the string
And OP is perpendicular to pa
So p is on a circle with diameter OA: (x-3) ^ 2 + y ^ 2 = 9
Then calculate the value range: consider the tangent between O line and x ^ 2 + y ^ 2-12x = 0
The tangent of point o on the circle x = 0
So the midpoint P trajectory of the chord:: (x-3) ^ 2 + y ^ 2 = 9, remove the origin

The equation of a circle is x2 + y2-6x-8x = 0. Through the origin of coordinates, make a string of length 8, and find the linear equation of the string

The original formula is: (x-3) & sup2; + (y-4) & sup2; = 25
Let the linear equation be: kx-y = 0
yes:
(|3k-4|)/√(k²+1)=√(5²-4²)
（3k-4）²=9k²+9
9k²-24k+16=9k²+9
7-24k=0
k=7/24
When x = 0
y²-8y=0
y1=0
y2=8

A plane π passes through the spherical center of the plane x ^ 2 + y ^ 2 + Z ^ 2 = 4x-2y-2z and is perpendicular to the straight line LX = 0, y + Z = 0
Find the projection of the intersection of the plane and the sphere on the xoy coordinate plane

x^2+y^2+z^2=4x-2y-2z
(x-2)^2+(y+1)^2+(z+1)^2=6
∵ the plane π is perpendicular to the line L x = 0, y + Z = 0
The positive direction of the plane π and the Y axis of the xoy coordinate plane is 135 and 186;
And o '(2, - 1, - 1)
The projection of the intersection of plane and sphere on the xoy coordinate plane
It is an ellipse whose major axis is parallel to the X axis. Its length is the diameter of the ball 2 √ 6
The short axis is parallel to the Y axis, and the length is 2 √ 6 * cos45 & # 186; = 2 √ 3
The center is (2, - 1,0)
The equation is (X-2) &# 178 / 6 + (y + 1) &# 178 / 3 = 1 and z = 0

Find the bisector equation of the angle between the plane x + 2y-2z + 6 = 0 and the plane 4x-y + 8z-8 = 0

The angular bisector must pass the intersection of plane 1: x + 2y-2z + 6 = 0 and plane 2: 4x-y + 8z-8 = 0
Let the equation of angular bisector be λ (x + 2y-2z + 6) + 4x-y + 8z-8 = (λ + 4) x + (2 λ - 1) y + (8-2 λ) Z + (6 λ - 8) = 0
Normal vector N1 of plane 1 = (1,2, - 2), normal vector N2 of plane 2 = (4, - 1,8), normal vector n of angular bisector = (λ + 4,2 λ - 1,8-2 λ)
Then according to the properties of angular bisector: there is (n, N1) = (n, N2), that is, the angle between N and N1 is equal to the angle between N and N2
Finally, the equation of angle bisector is 7x + 5Y + 2Z + 10 = 0