Given that circle a: (x + 3) 2 + y2 = 100, a certain point B (3, 0) in circle a, moving circle P passes through point B and is tangent to circle a, the trajectory equation of circle center P is obtained

Given that circle a: (x + 3) 2 + y2 = 100, a certain point B (3, 0) in circle a, moving circle P passes through point B and is tangent to circle a, the trajectory equation of circle center P is obtained

Let the moving circle center P (x, y), radius r, the center of ⊙ A is a (- 3, 0), radius 10, and because the moving circle passes through point B, r = Pb. If the moving circle P is inscribed with ⊙ a, then PA = 10-r = 10-pb, that is PA + Pb = 10 & nbsp; from (3) and (4), we get | PA + Pb | = 10 ＞ | ab | = 6, so the trajectory of point P is focused on a and B

Find two known circles C1: (x + 3) &# 178; + Y & # 178; = 1 and C2: (x + 3) &# 178; = 1_ 3) The orbit equation of the center of a moving circle which is both inscribed with y and Y

The center of C1 is a (- 3,0), radius R1 = 1
The center of C2 is B (3,0), and the radius is R2 = 3
Note that the center of the circle is C (x, y) and the radius is R. since C1 and C2 do not intersect, C can only contain C1 and C2
Then CA = r-r1 = R-1, CB = r-r2 = R-3
That is ca-cb = 2
This is the right half of the hyperbola,
2a=2,c=3
We obtain a = 1, C = 3, B & # 178; = C & # 178; - A & # 178; = 8
So the trajectory is: X & # 178; - Y & # 178 / 8 = 1, (x > 0)

1. It is known that the moving circle C passes through the point (- 3,0) and is tangent to the fixed circle B in the interior of the fixed circle B: (x-3) ^ 2 + y ^ 2 = 64

Let Center (a, b), radius R B, Center (3,0), radius 8 C be in B, and inscribed, so 8 > R center distance = 8-r, so (A-3) ^ 2 + (B-0) ^ 2 = (8-r) ^ 2 (1) circle is (x-a) ^ 2 + (y-b) ^ 2 = R ^ 2 passing through a, (- 3-A) ^ 2 + B ^ 2 = R ^ 2 (2) (1) - (2) - 12a = 64-16r, r = 4 + 3A / 4, so (- 3-A) ^ 2 + B ^ 2 = (4 + 3A /

Given that the moving circle P passes through the fixed point a (- 3,0) and is inscribed with the fixed circle B: (x-3) 2 + y2 = 64, then the locus of the center P of the moving circle is ()
A. Line segment B. straight line C. circle D. ellipse

As shown in the figure, let the moving circle P and the fixed circle B be inscribed at m, then the sum of the distances from the center P of the moving circle to two points, that is, the distance between the fixed point a (- 3,0) and the center B (3,0) of the fixed circle is exactly equal to the radius of the fixed circle, that is, | PA | + | Pb | = | PM | + | Pb | = | BM | = 8

Given the circle a: (x + 3) 2 + y2 = 4, fixed point (3,0), find the trajectory equation of the center P of the moving circle C which is tangent to the circle a

Let two points F1 and F2, the relation pf1-pf2 = 2 can be obtained from the problem, since the difference between the distance from the point to two fixed points is a certain value, satisfying the hyperbolic definition. Then a = 1, C = 3, then the square of B is equal to eight. Then you will understand that some problems are directly done with the definition. Of course, you can also directly formulate the relation, but it is more troublesome. At the same time, don't forget that x should be greater than 0

Given that the moving circle C passes through the fixed point a (- 5,0) and is tangent to the fixed circle B in the interior of the fixed circle B: (x-3) ^ 2 + y ^ 2 = 64, the trajectory equation of the center C of the moving circle is obtained

(- 5,0) is on circle B
So the trajectory is x-axis, between X ∈ (- 5,11)

If the quadratic function y = x2 + 0.5 coincides with the vertex of the image of y = - x2 + K, then the following conclusions are correct
A. The two function images have the same axis of symmetry B. the opening directions of the two function images are opposite
C. The equation - x2 + k = 0 has no real root C. The maximum value of quadratic function y = - x2 + k is K
(the answer seems to be more than one.)

Obviously abd is right
Only C error

The image of the quadratic function y = X2 - (k-1) x-3k-2 intersects with the X axis at a (a, 0) B (B, 0), and A2 + B2 = 17

Because the image of quadratic function y = x & sup2; - (k-1) x-3k-2 intersects with X axis at a (a, 0) B (B, 0), a and B are two real roots of X & sup2; - (k-1) x-3k-2 = 0, so a + B = k-1, ab = - 3K-2, △ = (k-1) & sup2; + 4 (3K + 2) ≥ 0, because a & sup2; + B & sup2; = 17, that is: A & sup2; + B & sup2; = (a + b)

It is proved that the image of quadratic function y = X2 - (K + 2) x + K-3 and X-axis always have two common points

Discriminant = [- (K + 2)] & sup2; - 4 (K-3)
=K²+4K+4-4K+12
=K²+16
K²>=0
So K & sup2; + 16 > = 16 > 0
The discriminant is greater than 0, so there must be two intersections with X axis

The image of quadratic function y = the square of a (X-H) + K passes through points (- 2,0) and (4,0). Try to determine the value of H

From the passing points (- 2,0) and (4,0), we can see that the axis of symmetry is [4 - (- 2)] / 2 = 3, so - (- H) = 3, then H = 3