Let the center of the circle x ^ 2 + y ^ 2-4x + 2y-11 = 0 be a and the point p be on the circle, then the trajectory equation of the midpoint m of PA is

Let the center of the circle x ^ 2 + y ^ 2-4x + 2y-11 = 0 be a and the point p be on the circle, then the trajectory equation of the midpoint m of PA is

x^2+y^2-4x+2y-11=0
That is, (X-2) &# 178; + (y + 1) &# 178; = 4 & # 178;
The trajectory of PA key point m is a circle with a as the center and 1 / 2 of the original circle radius as the radius. The equation is
(x-2)²+(y+1)²=2²=4

If the circle x2 + y2-ax + 2Y + 1 = 0 and the circle x2 + y2 = 1 are symmetric with respect to the straight line y = X-1, and the circle P passing through the point C (- A, a) is tangent to the Y axis, then the trajectory equation of the center P is ()
A. y2-4x+4y+8=0B. y2-2x-2y+2=0C. y2+4x-4y+8=0D. y2-2x-y-1=0

The center (A2, − 1) of circle x2 + y2-ax + 2Y + 1 = 0, because circle x2 + y2-ax + 2Y + 1 = 0 and circle x2 + y2 = 1 are symmetric with respect to the straight line y = X-1, so (A4, − 12) satisfies the equation of straight line y = X-1, the solution is a = 2, the circle P passing through point C (- 2,2) is tangent to the Y axis, and the coordinates of the center P are (x, y), so (x + 2) 2 + (Y − 2) 2 = | x | & nbsp; the solution is: Y2 + 4x-4y + 8 = 0, so choose C

Given that circle a (x + 2) ^ 2 + y ^ 2 = 16, circle B (X-2) ^ 2 + y ^ 2 = 4, moving circle C is inscribed with circle a and circumscribed with circle B, then the center of moving circle is
The answer to the trajectory equation is x ^ 2 / 9 + y ^ 2 / 5 = 1, where - 3 ≤ x ＜ 3 / 2. How can we get the range of X? How can we get the detailed solution

Given circle a: (x + 2) &# 178; + Y & # 178; = 16; circle B: (X-2) &# 178; + Y & # 178; = 4; moving circle C is inscribed with circle a and circumscribed with circle B, the trajectory equation of the center of the moving circle is obtained
The radius of circle a (- 2,0) is 4; the radius of circle B (2,0) is 2;
Let the coordinates of the moving center C be (x, y); let C and a be inscribed at D, and B be circumscribed at e; then a, C and D are in a straight line
And ∣ ad ∣ - ∣ AC ∣ = ∣ BC ∣ - ∣ be ∣ = radius of moving circle C
Where ∣ ad ∣ = 4, ∣ AC ∣ = √ [(x + 2) &# 178; + Y & # 178;]; ∣ BC ∣ = √ [(X-2) &# 178; + Y & # 178;], ∣ be ∣ = 2; therefore, the equation is as follows:
4 - √ [(x + 2) & # 178; + Y & # 178;] = √ [(X-2) & # 178; + Y & # 178;] - 2, that is, 6 - √ [(x + 2) & # 178; + Y & # 178;] = √ [(X-2) & # 178; + Y & # 178;]. (1)
Square both sides of (1) to 36-12 √ [(x + 2) & # 178; + Y & # 178;] + (x + 2) & # 178; + Y & # 178; = (X-2) & # 178; + Y & # 178;];;
The expansion is reduced to 9 + 2x = 3 √ [(x + 2) &# 178; + Y & # 178;];
Square again and get 81 + 36x + 4x & # 178; = 9 (X & # 178; + 4x + 4 + Y & # 178;)
To simplify, the trajectory equation of the center C of the moving garden is: 5x & # 178; + 9y & # 178; = 45, and the standard form is X & # 178 / 9 + Y & # 178 / 5 = 1
Domain of definition: from 16 - (x + 2) &# 178; = 4 - (X-2) &# 178;, the abscissa of the intersection of circles a and B is x = 3 / 2; the leftmost position of the center C of the moving circle is
X = - 3, so the definition domain of trajectory equation is - 3 ≤ x ≤ 3 / 2

Given that the moving circle C and the fixed circle C1: x ^ 2 + (y-4) ^ 2 = 64 inscribed, and the fixed circle C2: x ^ 2 + (y + 4) ^ 2 = 4 circumscribed, let C (x, y), then 25X ^ 2 + 9y ^ 2 =?

Let the center of circle C be (a, b) and the radius be r
If circle C and circle C1 are inscribed, then
a^2＋（b－4)^2=(8-r)^2
If circle C and circle C2 are circumscribed, then
a^2＋（b+4)^2=(2+r)^2
subtract
r＝3＋4b/5
Substituting a ^ 2 + (B + 4) ^ 2 = (2 + R) ^ 2, we get
25a^2＋9b^2－225＝0
Changing a, B to x, y is the best way
25x^2＋9y^2－225＝0