Given that circle C1: (x + 3) square + y square = 9 and circle C2: (x-3) square + y square = 81, moving circle m and Circle C1 is circumscribed and circle C2 is inscribed. The trajectory equation of moving circle center m is obtained

Given that circle C1: (x + 3) square + y square = 9 and circle C2: (x-3) square + y square = 81, moving circle m and Circle C1 is circumscribed and circle C2 is inscribed. The trajectory equation of moving circle center m is obtained

From the circular equation, C1: (- 3,0) C2: (3,0)
From the approximate image, it can be concluded that the trajectory function of the moving circle center m is the inverse function of the quadratic function
We get x = ay & # 178; + by + C
From the image, the vertex of the function is (6,0) [the vertex is the intersection of the image and the X axis]
So B = 0, C = 6
Take a point n on the Y axis so that n satisfies the M condition
Connecting c1n and C2n
Let the radius of circle n be r
From right triangle c2no
ON²=（9-R）²-9
From right triangle c1no
（9-R）²-9+9=（R+3）²
The solution is r = 3
So on = 3 √ 3
So n: (0,3 √ 3)
Bring in function
Get a = - 2-9
So the trajectory equation (function) of the moving circle center M: x = - 2, 9y + 6 (- 6 ＜ x ＜ 6) is changed to y = √ (- 18x + 108), 2 (- 6 ＜ x ＜ 6)
Happy new year, by the way!
It is known that circle C1: x-square + y-square-2ay + a-square-1 = 0 and circle C2: (x-3) square + (y + 2) square = 16 circumscribed
(1) Find the value of the real number a (2) if a > 0, find the equation of the straight line L passing through the point P (- 1,4) and tangent to the circle C1, and do it before 12:30
Circle C1: x ^ 2 + (Y-A) ^ 2 = 1, center C1 is (0, a), radius R1 = 1, circle C2: (x-3) ^ 2 + (y + 2) ^ 2 = 16, center C2 is (3, - 2), radius R2 = 4, (1), C1C2 = √ [(3-0) ^ 2 + (- 2-A) ^ 2] = R1 + R2 = 5, --- A ^ 2 + 4a-12 = (A-2) (a + 6) = 0, --- a = 2, or a = - 6, (2), a > 0, then a = 2, --
Circle C1: x ^ 2 + y ^ 2-2ay + A ^ 2-1 = 0 circle C2: (x-3) ^ 2 + (y + 2) ^ 2 = 16
Circle C1: x ^ 2 + (Y-A) ^ 2 = 1 circle C2: (x-3) ^ 2 + (y + 2) ^ 2 = 16
When two circles are circumscribed, the center distance is equal to the sum of their radii
The standard equation of two circles gives the radius of the center of circle C1 (0, a) 1, the center of circle C2 (3, - 2) 4
Circle C1: (x-4) ^ 2 + y ^ 2 = 169, circle C2: (x + 4) ^ 2 + y ^ 2 = 9, moving circle is inside circle C1 and inscribed with circle C1, and circle
Circle C1: (x-4) ^ 2 + y ^ 2 = 169, circle C2: (x + 4) ^ 2 + y ^ 2 = 9, the moving circle is inside circle C1 and is tangent to circle C1, and is tangent to circle C2
1 circle is tangent to the left of the first circle and the left of the second circle is 2 units long. In this case, the circle takes (- 8,0) as the center and 1 as the radius (x + 8) ^ 2 + y ^ 2 = 1
Circle 2 is tangent to the right of the second circle on the right of the first circle, and the distance is 18 units long. In this case, the circle takes (8,0) as the center and 9 as the radius (X-8) ^ 2 + y ^ 2 = 81
(x + 8) ^ 2 + y ^ 2 = 1 or (X-8) ^ 2 + y ^ 2 = 81
Center C (x, y), radius r, circle C and C1 inscribed | CC1 | = 13-r, circle C and C2 circumscribed, | CC2 | = R + 3 | CC1 | + | CC2 | = 16, C1 (4,0) C2 (- 4,0) ellipse C = 4, a = 8, the equation is x ^ 2 / 64 + y ^ 2 / 48 = 1
It is proved that the function f (x) = x − 2x + 1 is an increasing function on (- 1, + ∞) by monotonicity definition
∵ f (x) = x − 2x + 1 = 1-3x + 1, let x1, X2 ∈ (- 1, + ∞), and X1 < X2, ∵ f (x1) - f (x2) = 1-3x1 + 1-1 + 3x2 + 1 = 3 (x1 − x2) (x1 + 1) (x2 + 1), because - 1 < X1 < X2, so x1-x2 < 0, X1 + 1 > 0, X2 + 1 > 0, so f (x1) - f (x2) < 0, that is, f (x1) < f
It is proved that the square-2x of F (x) = x is an increasing function on {x | x > = 1} by the definition of function monotonicity
It is proved that any two points x1, X2, X1 > x2 > = 1
f(x1)-f(x2)=(x1^2-2x1)-(x2^2-2x2)
=(x1+x2)(x1-x2)-2(x1-x2)
=(x1-x2)(x1+x2-2)
Because x1-x2 > 0, X1 + x2 > 2, that is, X1 + x2-2 > 0
Therefore, f (x1) - f (x2) > 0
That is: F (x1) > F (x2)
Therefore, the function is an increasing function on [1, + infinity]
Given the function f (x) = (2x + 1) / (x + 1) ①, judge the monotonicity of the function on [1, + ∞), and prove it by definition
Known function f (x) = (2x + 1) / (x)
+1)
① The monotonicity of [1 + ∞) is proved
② Find the maximum and minimum of the function on [1,4] (write the process)
①∵f(x)=(2x+2-1)/(x+1)=2-1/(x+1)
When ∵ x ∈ [1, + ∞), - 1 / (x + 1) monotonically increases
The monotonic increase of ∧ f (x)
② From (1), we obtain that f (x) increases monotonically on [1, + ∞)
When x ∈ [1,4], the minimum value is 3 / 2 when x = 1, and the maximum value is 9 / 5 when x = 4
It is proved that the function y = sinx-x monotonically decreases
Derivation of Y
y'=cosx-1
Because cosx is always less than 1, so y '
It is proved that the function y = x-sinx increases monotonically
y'=(x-sinx)'=1-cosx
-1≤cosx≤1
1-cosx≥0
y'≥0
The function increases monotonically
The function whose derivative at point = 0 equals 0 is a.y = SiNx b.y = X-1 C.Y = e ^ x-x D,
When x = 0, a Y '= cosx = 1, b y' = 1, C Y '= e ^ X-1 = 0, d y' = 2x-1 = - 1
So choose C
Using derivative knowledge to prove inequality and calculus
It is proved that when x > 0, there are (1 + x) ㏑ & # 178; (1 + x) > X & # 178;
Proof: Let f (x) = (1 + x) ln & # 178; (1 + x) - X & # 178;,
Then f (x) is continuously differentiable in (0, + ∞)
F '(x) = ln & # 178; (1 + x) + 2ln (1 + x) - 2x, such that G (x) = f' (x), where,
Then G '(x) = [2ln (1 + x)] / (1 + x) + 2 / (1 + x) - 2 = [2ln (1 + x) + 2 - (2 + 2x)] / (1 + x) = 2 [ln (1 + x) - x] / (1 + x),
Let H (x) = ln (1 + x) - x,
Then when x ∈ (0, + ∞), H '(x) = - X / (x + 1)