Given that the moving circle m is tangent to the straight line y = 2 and circumscribed with the fixed circle C: x 2 + (y + 3) 2 = 1, the trajectory equation of the moving circle m is obtained

Given that the moving circle m is tangent to the straight line y = 2 and circumscribed with the fixed circle C: x 2 + (y + 3) 2 = 1, the trajectory equation of the moving circle m is obtained

The moving circle m is tangent to the straight line y = 2 and circumscribed to the fixed circle C: x2 + (y + 3) 2 = 1
The distance from the moving point m to C (0, - 3) is equal to the distance to the straight line y = 3
According to the definition of parabola, the locus of point m is a parabola with C (0, - 3) as the focus and the straight line y = 3 as the guide line
Therefore, the trajectory equation of M is x2 = - 12Y

The trajectory equation of the center of a circle passing through point a (2,0) and inscribed with circle x2 + 4x + y2-32 = 0 is obtained

Let the coordinates of the center of the moving circle be (x, y), from x2 + 4x + y2-32 = 0, we can get: (x + 2) 2 + y2 = 36,
The center coordinate of circle x2 + 4x + y2-32 = 0 is (- 2, 0) and the radius is 6
∵ the moving circle passes through point a (2,0) and is inscribed with circle x2 + 4x + y2-32 = 0,
Qi
(x−2)2+y2＝6−
(x+2)2+y2，
The square of both sides is: x2 − 4x + 4 + y2 = 36 − 12
(x+2)2+y2+x2+4x+4+y2，
That is 3
(x+2)2+y2＝9+2x．
The result is: 5x2 + 9y2 = 45
That is, x2
9+y2
5＝1．

The moving circle and the fixed circle x ^ 2 + y ^ 2 + 4y-32 = 0 are inscribed and pass through the fixed point a (0,2), and the trajectory equation of the center P of the moving circle is solved

x²+(y+2)²=36
Center B (0, - 2)
Radius 6
Let the mobilization radius be r
Center C (x, y)
Then r = AC
Inscribed BC = 6-r
So AC + BC = 6
So it's an ellipse. AB is the intersection
Then C = 2,2a = 6, a = 3
b²=9-4=5
So x 2 / 5 + y 2 / 9 = 1

The trajectory equation of the center of a circle tangent to the circle x2 + y2-4x = 0 and tangent to the y-axis is () A. y2=8x B. Y2 = 8x (x > 0) and y = 0 C. y2=8x（x＞0） D. Y2 = 8x (x > 0) and y = 0 (x < 0)

Let the center of the circle tangent to the Y axis and circumscribed to the circle C: x2 + y2-4x = 0 be p (x, y), and the radius is r,
be
(x−2)2+y2=|x|+2，
If x > 0, then y2 = 8x; if x < 0, then y = 0;
Therefore, D

Given the fixed circle C1: x ^ 2 + y ^ 2 + 4x = 0, the fixed circle C2: x ^ 2 + y ^ 2-4x-60 = 0, the moving circle m and the fixed circle C1 circumscribed and the circle C2 inscribed, the trajectory equation of the moving circle center m is obtained

Let m (x, y) be the center of a moving circle
C1：(x+2)²+y²=4→C1(-2,0),r1=2
C2：(x-2)²+y²=64→C2(2,0),r2=8
And C1 circumscribed →| MC1 | = R1 + R
And C2 endotangent →| MC2 | = | r2-r|
① If R2 > R, then | MC2 | = r2-r
∴|MC1|+|MC2|=r1+r2=10
From the definition of ellipse, we know that the trajectory of M is an ellipse, and the focus is C1, C2, and the focal length C = 2
2A = 10 leads to a = 5
∴b²=25-4=21
The M trajectory equation is: x 2 / 25 + y 2 / 21 = 1
②r2∴|MC1|-|MC2|=r1+r-r+r2=10,
∵|C1C2|=4
∴|MC1|-|MC2|>|C1C2|
According to the difference between the two sides of the triangle less than the third side, we know that M has no solution
So the trajectory equation of M is: x 2 / 25 + y 2 / 21 = 1

The trajectory equation of the center of a circle tangent to the X axis and circumscribed to the circle x2 + y2 = 1 is () A. x2=2y+1 B. x2=-2y+1 C. x2=2|y|+1 D. x2=2y-1

Let the center of the circle tangent to the X axis and circumscribed to the circle C: x2 + y2 = 0 be p (x, y), and the radius is r,
be
x2+y2=r+1，|y|=r，
Qi
x2+y2=|y|+1，
Square x 2 = 2|y| + 1
Therefore, C

Trajectory equation of the center of a moving circle tangent to the Y axis and circumscribed to the circle x ^ 2 + y ^ 2-6x + 8 = 0

It is known that the circle equation is (x-3) ^ 2 + y ^ 2 = 1, and the center of the circle is C (3,0)
Let the center of the circle be m (x, y)
Then │ x │ = │ MC │ - 1, i.e., │ MC │ - x │ = 1
The difference between the distance to the fixed point C and the fixed straight line (Y axis) is equal to the constant 1
The locus of the point is the right half branch of the parabola
p/2=3,p=6
The equation is y ^ 2 = 12x

Find the equation of circle passing through the intersection point and point (2,1) of two circles x ^ 2 + y ^ 2 = 1, x ^ 2 + y ^ 2-4x + 4Y + 1 = 0 Find the equation of circle passing through the intersection point and point (2,1) of two circles x ^ 2 + y ^ 2 = 1, x ^ 2 + y ^ 2-4x-4y + 1 = 0

X ^ 2 + y ^ 2-1 + K (x ^ 2 + y ^ 2-4x-4y + 1) = 0, take x = 2, y = 1 to get the value of K

Find the equation of the circle with the point (3,1) of the intersection point of two circles x ^ 2 + y ^ 2-x-y-2 = 0 and x ^ 2 + y ^ 2 + 4x-4y-8 = 0

Let the new equation be as follows:
x^2+y^2-x-y-2+k(x^2+y^2+4x-4y-8)=0 （1）
Substituting the point (3,1), k = - 0.4 is obtained
By substituting (1), the circular equation is obtained as follows:
3x^2+3y^2-13x+3y+6=0

Find the equation of the circle with the point (3,1) of the intersection point of two circles x ^ 2 + y ^ 2-x-y-2 = 0 and x ^ 2 + y ^ 2 + 4x-4y-8 = 0 X ^ 2 + y ^ 2-x-y-2 + λ (x ^ 2 + y ^ 2 + 4x-4y-8) = 0, I know how to list this, but I want to ask why this is listed

The iron armour is still in: what the landlord said: x? + y? - x-y-2 + λ (x? + y? + 4x-4y-8) = 0 is a circular system equation: circle C1: x? + y? + D1X + e1y + F1 = 0? Circle C2: x? + y? + d2x + e2y + F2 = 0 if two circles intersect, then the circular system equation passing through the intersection point is: X -