Given that the circle C: x 2 + y 2 = 5, the straight line L: ax-y-2a = 0 and circle C intersect at two points AB, the trajectory equation of the midpoint of chord AB is obtained

The solution is determined by the line L: ax-y-2a = 0
L: a (X-2) - y = 0
When x = 2, y = 0
Know the constant crossing point m (2,0) of the straight line L
From 2 ^ 2 + 0 ^ 2 = 4 < 5
It is known that m (2,0) is in the circle C: x 2 + y 2 = 5
Let the midpoint of chord AB be t (x, y)
Then the image is combined to know
The line to is perpendicular to the line TM
That is, ktoktm = - 1
(y) × (0) / (0) = (0)
That is y ^ 2 / (x-0) (X-2) = - 1
That is, y ^ 2 = - (x ^ 2-2x)
So the locus equation of the midpoint of string ab
y^2+X^2-2x=0

If the square of the circle x + the square of y = 4, a point [1.0] triggers the chord AB, then the locus of the midpoint D of ab

Let the slope of chord AB be K A (x1, Y1) B (X2, Y2)
The equation for string AB is
y=k(x-1)
Substituting 4x ^ 2 + 9y ^ 2 = 36
4x^2+9k^2(x-1)^2=36
x1+x2=18k^2/(9k^2+4)
Abscissa of midpoint M = 9K ^ 2 / (9K ^ 2 + 4)
y1+y2=-8k/(9k^2+4)
The ordinate of the midpoint of ^ 4K / (4k-2m)
x/y=-9k/4
K = - 4x / 9y is obtained by substituting x = 9K ^ 2 / (9K ^ 2 + 4)
4x^2+4x+9y^2=0
4(x+1/2)^+9y^2=1

Through a point P (1,0) in the ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1, the chord AB is induced and the trajectory equation of the midpoint m of the chord is obtained

Go over the circle O: x ^ 2 + y ^ 2 = 16. Make a straight line intersect the circle O at two points ab. find the locus of the midpoint C of the chord ab If the elimination parameter method is used: when the slope of line AB does not exist, the midpoint of chord AB is C (2,0) When the slope of line AB exists, let it be K, and the equation of line AB is y = K (X-2) - 6 From y = kx-2k-6 x^2+y^2=16 (k ^ 2 + 1) x ^ 2 - (4K ^ 2 + 12K) x + 4K ^ 2 + 24K = 0 From △ > 0, K belongs to (- ∝, 0) ∪ (3 / 4, + ∞) Let C (x0, Y0) then x0 = (x1 + x2) / 2 = (4K ^ 2 + 12K) / (2k ^ 2 + 2) y0=(y1+y2)/2=k(x0-2)-6 I won't be able to cancel the parameters here?

When the slope of line AB exists, let it be K, and the equation of line AB is y = K (X-2) - 6
From y = kx-2k-6
x^2+y^2=16
(k ^ 2 + 1) x ^ 2 - (4K ^ 2 + 12K) x + 4K ^ 2 + 24K = 0
Let a (x1, Y1), B (X2, Y2)
Let C (x0, Y0) then x0 = (x1 + x2) / 2 = (4K ^ 2 + 12K) / (2k ^ 2 + 2)
A. B on a straight line, it satisfies the linear equation
y1=k(x1-2)-6
y2=k(x2-2)-6
y1+y2=k(x1+x2-4)-12
=k[(4k^2+12k)/(1+k^2)-4]-12
=k[(4k^2+12k)-4-4k^2]/(1+k^2)-12
=k(12k-4)/(1+k^2)-12
=[k(12k-4)-12(1+k^2)]/(1+k^2)
=(-4k-12)/(1+k^2)
y0=(y1+y2)/2=(-2k-6)/(1+k^2) (1)
x0=(2k^2+6k)/(1+k^2)=k(2k+6)/(1+k^2) (2)
(2)/(1)
x0/y0=-k
Because: Y0 = K (x0-2) - 6
=(-x0/y0)(x0-2)-6
Y0 ^ 2 = - x0 ^ 2 + 2x0-6y0, replace x0, Y0 with X, y is the trajectory equation of midpoint C,

Go through the circle x ^ 2 + y ^ 2 = 5, make a straight line P (4,0) outside the circle and intersect the circle at two points a and B. find the locus of the midpoint m of the chord ab Find the locus equation of midpoint m of string ab

Using the vertical diameter theorem, then om ⊥ AB is om ⊥ PM. If M (x, y) is set, then om = (x, y), PM = (x-4, y)  OM.PM=0 In other words, X (x-4) + y? = 0  x? + y? - 4x = 0 ﹥ X-2)  + y? = 4. It is noted that the midpoint of the chord needs to be inside the circle

Circle equation for x ^ 2 + y ^ 2-6x-8y = 0, through the origin of coordinates as a chord of length 6, find the line equation of the chord. Urgent!

The circle is reduced to the standard equation: (x-3) 2 + (y-4) 2 = 25,
Half chord length, radius and the distance from the center of a circle to a straight line form a right triangle
The distance from the center of the circle to the straight line is 4,
Let the linear equation be y = KX and the coordinates of the circle center be (3,4),
From the distance formula of point to straight line, K-4 | / √ (K | + 1) = 4, k = 0 or - 24 / 7
So the linear equation is y = 0 or 7Y + 24x = 0
(note that for straight lines and circles, the combination of number and shape should be used as much as possible to simplify the calculation.)

The equation of the circle is x2 + y2-6x-8y = 0. If a chord of length 8 is made through the origin of coordinates, then the equation of the line where the chord is located is______ (general equation of the result written as a straight line)

X2 + y2-6x-8y = 0, i.e. (x-3) 2 + (y-4) 2 = 25. If the slope exists, let the straight line be y = KX
∵ the radius of the circle is 5, and the distance from the center m (3, 4) to the straight line is 3, ᙽ d = | 3K − 4|
k2+1=3，
∴9k2-24k+16=9（k2+1），∴k=7
24. The straight line is y = 7
24x；
When the slope does not exist, if the straight line is x = 0, verify that its chord length is 8, so x = 0 is also the line to be calculated. Therefore, the straight line is: x = 0 or 7x-24y = 0
So the answer is: x = 0 or 7x-24y = 0

The equation of the circle is x * 2 (square) + y * 2 (square) - 6x-8y = 0. Make a chord of length 8 through the coordinate origin, and find the line equation of the chord

The chord length of the circle on the y-axis is 8, which is x = 0;
The length of the symmetrical line of the chord between the center of the circle and the origin is 8,
From the slope product = - 1 and the distance from the point to the line is equal
(0,8) on the symmetry point of y = 4 / 3x (7.68,2.24)
So the linear equation is y = 7 / 24x

The equation of the circle is x2 + y2-6x-8y = 0. If a chord of length 8 is made through the origin of coordinates, then the equation of the line where the chord is located is______ (general equation of the result written as a straight line)

The equation of the circle is x ^ 2 + y ^ 2 + 8x-6y = 0. Find the chord with length of 8 through the coordinate origin, and find the line where the chord is

For a straight line passing through the coordinate origin, y = KX chord length = 8, half of the chord = 4 radius = 5, so the chord center distance = under the root sign (25-16) = 3 the distance from the center of the circle to the straight line = | 4K + 3 | / under the root sign (K | 1) = chord center distance = 3 (4K + 3)

Given that the circle C: x 2 + y 2 = 5, the straight line L: ax-y-2a = 0 and circle C intersect at two points AB, the trajectory equation of the midpoint of chord AB is obtained