Through the origin, make the secant of the square of the sum of circles x + 1 plus the square of Y minus 2, intersect the circle at two points a and B, and find the trajectory equation of the midpoint of chord AB and m

Through the origin, make the secant of the square of the sum of circles x + 1 plus the square of Y minus 2, intersect the circle at two points a and B, and find the trajectory equation of the midpoint of chord AB and m

The center of the circle is a (- 1,2), and the angle amo is a right angle,
So the trajectory of M is a circle with the diameter of Ao,
The center of the circle is the midpoint of A0 (- 1 / 2,1), and the radius is AO / 2 = √ 5 / 2, so the trajectory equation is
Square of (x + 1 / 2) + square of (Y-1) = 5 / 4

The path equation of the middle point m of the chord AB is obtained by the intersection of the secant of the leading circle from a point P (5.12) outside the circle x + y = 9 at two points a and B

Y-12 = K (X-5)
We get y = K (X-5) + 12
Circle x square + y square = 9, a point P (5,12) leads a straight line to intersect a and B
x^2+y^2=9
x^2+[k(x-5)+12]^2=9
Two simultaneous equations: x ^ 2 + K ^ 2 * x ^ 2-10k ^ 2 * x + 24kx + 25K ^ 2 + 144-9 = 0
(k^2+1)x^2-(10k^2-24k)x+25k^2+135=0
When tangent, diata = (10K ^ 2-24k) ^ 2-4 * (k ^ 2 + 1) * (25K ^ 2 + 135) = 0
Two values of K are obtained. These two values are the two maximum and minimum values of K (they can't be equal to each other because they intersect, there are two intersections, and they can't be tangent to one intersection)
The midpoint of the two intersections is: x = (x1 + x2) / 2 = (5K ^ 2-12k) / (k ^ 2 + 1) a
y=k(x-5)+12=kx-5k+12
y=(y1+y2)/2=k*(x1+x2)/2-5k+12 B
By eliminating K from the two equations a and B, the trajectory equation of the midpoint m of the chord AB is obtained, and the value range of X can be obtained from the value range of K

It is known that AB is a moving chord whose square of circle C: x plus the square of Y is equal to 25, and the absolute value of AB is equal to six, then the trajectory equation of the midpoint P of AB is
1: π 2 / 6: π 3 / 4: π 4 / 3: π 2 / 2

Connect OP,
According to the vertical diameter theorem op ⊥ AB,
AP=1/2AB=3,
OA=5,∴OB=√(OA^2-AP^2)=4,
The trajectory of P is a circle with o as the center and 4 as the radius
Trajectory equation:
X^2+Y^2=16.

Find the trajectory equation of the midpoint of the chord of ellipse x ^ 2 / 4 + y ^ 2 = 1 made by point m (1,0)

Let the midpoint coordinates be (x, y), the endpoint coordinates be (x1, Y1), (X2, Y2) respectively, and substitute them into the elliptic equation, and the subtraction result is: (x1-x2) * (x1 + x2) / 4 + (y1-y2) (Y1 + Y2) = 0. Divide them by (x1-x2) on both sides, and substitute them with the midpoint formula x = (x1 + x2) / 2; y = (Y1 + Y2) / 2 to get: X / 2 + [(y1-y2) / (x1-x2)] (2

Find the locus equation of the middle point of the chord of the ellipse x ^ 2 / 4 + y ^ 2 = 1 made by point m (1,0)

Let the line passing through point m be x = my + 1
Substituting into the equation x & sup2; + 4Y & sup2; = 4
m²y²+2my+1+4y²=4
(m²+4)y²+2my-3=0
y1+y2=-2m/(m²+4)
x1+x2=m(y1+y2)+2=-2m²/(m²+4)+2=8/(m²+4)
Let P (x, y) be the midpoint of the chord
x=(x1+x2)/2=4/(m²+4)
y=(y1+y2)/2=-m/(m²+4)
y/x=-m/4
x=my+1
m=(x-1)/y
therefore
y/x=(1-x)/4y
x²-x+4y²=0
(x-1/2)²+4y²=1/4

Practice of absolute value equation of one degree with one variable
Let a and B be rational numbers, and | a | > 0. The equation | x-a | - B | = 3 has three unequal solutions. Find the value of B
The process

The original equation can be reduced to | x-a | - B = ± 3, that is | x-a | = B ± 3
1. If B-3 and B + 3 are both greater than 0, the solution of the equation is
a±（b-3）,a±（b+3）,
When B ≠ 0, the four solutions are different; when B = 0, the original equation has only two solutions
2. If one of B-3 and B + 3 is 0, then first let B-3 = 0 and the equation has three solutions x = a, x = a + 6 and x = a-6; then let B + 3 = 0 and the equation has only one solution x = a
3. If B-3 and B + 3 are less than 0, then the number of solutions is less than 3
In conclusion, when B = 3, the equation has three unequal solutions

Simple absolute value, one variable, one equation, urgent!

|x-2|=3

How to do linear equation with absolute value sign?

The unknown is X
Classification discussion:
If x is a positive number, then x is equal to X
If x is negative, then x is equal to - X
If x is equal to 0, then x is equal to 0, x, - X

Given that (M + 3) x ^ / M / - 2 = 18 [/ / is the sign of absolute value] is a linear equation of one variable with respect to x, then M = ()

According to the meaning of the title:
|m|=1
∴m=±1

(M's square-4) x's Square - (m-2) x + 8 = 0 is a linear equation of one variable with respect to X. its solution is n. find the value of the absolute value of the equation m times Y-2 with respect to y = n

The square of (M-4) x - (m-2) x + 8 = 0 is a linear equation of one variable about X
therefore
m²-4=0
m-2≠0
therefore
m=-2
The equation is 4x + 8 = 0
x=-2=n
therefore
The absolute value of M times Y-2 = n
-2×|y-2|=-2
Y-2 = 1 or Y-2 = - 1
Y = 3 or 1