Given the circle C: (x + 1) 2 + y2 = 25 and point a (1, 0), q is a point on the circle, and the vertical bisector of AQ intersects CQ at m, then the trajectory equation of point m is______ ．

Given the circle C: (x + 1) 2 + y2 = 25 and point a (1, 0), q is a point on the circle, and the vertical bisector of AQ intersects CQ at m, then the trajectory equation of point m is______ ．

From the equation of the circle, we can see that the center of the circle is C (- 1,0), the radius is equal to 5, let the coordinates of the point m be (x, y), ∵ AQ's vertical bisector intersects CQ at m, | Ma | = | MQ |, and | MQ | + | MC | = radius 5, | MC | + | Ma | = 5 ＞| AC |. According to the definition of ellipse, the locus of the point m is an ellipse with a and C as the focus, and 2A = 5, C = 1, | B = 212, so the elliptic equation is x2254 + & nbsp; Y2214 = 1, that is 4x225 + 4y221 = 1, so the answer is 4x225 + 4y221 = 1

Given that the coordinates of the endpoint B of the line AB are (4,3), the endpoint a moves on the circle (x + 1) (x + 1) + YY = 4, the locus of the midpoint m of the line AB is obtained
I can't play square, can I see it clearly?
The trajectory of point m is a circle, and the equation of circle is (x-1.5) ^ + (y-1.5) = 1

(x+1)^2+y^2=4
Parameter equation:
x=2cost-1,y=2sint
A(2cost-1,2sint),B(4,3)
Midpoint m (x, y)
x=(2cost-1+4)/2=(2cost+3)/2
y=(2sint+3)/2
,cost=(x-3/2),sint=(y-3/2)
cost^2+sint^2=(x-3/2)^2+(y-3/2)^2=1
The trajectory of point m is a circle,
The equation of circle is (x-1.5) ^ + (y-1.5) = 1

X / (x-1) = (Y & # 178; + 4x-2) / (Y & # 178; + 4x-1), then y & # 178; + 4Y + X is

Let y ^ 2 + 4Y + x = t
Y ^ 2 + 4Y = t-x is substituted into the equation
x/(x-1)=(t-x-2)/(t-x-1)
tx-x^2-x=tx-x^2-2x-t+x+2
(t-1)x-x^2=-x^2+(t-2+1)x+2-t
(t-1-t+2-1)x=2-t
0=2-t
t=2
That is: y ^ 2 + 4Y + x = 2

Solve the equations 4x21 + 4x2 = y4y21 + 4y2 = z4x21 + 4x2 = X

According to the meaning of the problem, from equation (1) and (3), we get: x = y, and ∵ x = y, ∵ y = z = x, ∵ 4x21 + 4x2 = X. the solution of the equation is: x = 0 or 12, ∵ the solution of the original equation is x = y = z = 12 or 0