It is known that circle C passes through point a (- 2,0) and point (0,2), and the center of circle is on the straight line y = x, and there is a straight line L = KX + 1 intersecting with circle C at two points P and Q It is known that circle C passes through point a (- 2,0) and point (0,2), and the center of circle is on the straight line y = x, and there is a straight line L = KX + 1 intersecting with circle C at P and Q. The equation of circle C is obtained

It is known that circle C passes through point a (- 2,0) and point (0,2), and the center of circle is on the straight line y = x, and there is a straight line L = KX + 1 intersecting with circle C at two points P and Q It is known that circle C passes through point a (- 2,0) and point (0,2), and the center of circle is on the straight line y = x, and there is a straight line L = KX + 1 intersecting with circle C at P and Q. The equation of circle C is obtained

If the center of the circle is in the straight line y = x, then let the coordinates of the center of the circle be (x, x) and there is the distance between two points. The formula knows that (X-2) ^ 2 + x ^ 2 = (x + 2) ^ 2 + x ^ 2, x = 0, y = 0, the coordinates of the center of the circle are (0,0), l = KX + 1 and circle C intersect at P and Q points, which is redundant information. Therefore, the equation of C is x ^ 2 + y ^ 2 = 4

Given circle C: X & # 178;; + Y & # 178;; - 2x + 4y-4 = 0, a straight line L with slope equal to 1 intersects circle C at points a and B, (1) find the longest time of chord ab
Given circle C: X & # 178;; + Y & # 178;; - 2x + 4y-4 = 0, a straight line L with slope equal to 1 intersects circle C at points a and B,
If ∠ AOB is an obtuse angle (where o is the origin of the coordinate), calculate the range of intercept of the line L on the y-axis

Circle C: (x-1) ^ 2 + (y + 2) ^ 2 = 9, C (1, - 2), r = 3
Let the intercept of L on the Y axis be C, and the equation be y = x + C, X - y + C = 0
The distance between C and l d = | 1 + 2 + C | / √ 2 = | C + 3 | / √ 2
Radius of circle with ab as diameter r = √ (R ^ 2 - D ^ 2) = √ [9 - (c + 3) ^ 2 / 2]
The origin o is in the circle with ab as the diameter, and the distance d '= | 0 - 0 + C | / √ 2 = | C | / √ 2
d' < R
|c|/√2 < √[9 -(c+3)^2/2]
c^2/2 < 9 - (c+3)^2/2
2c^2 + 6c -9 < 0
(-3 - 3√3)/2 < c < (-3 + 3√3)/2

Given the circle C: X & # 178; + Y & # 178; - 2x + 4y-4 = 0, ask if it is a straight line L with a slope of 1, and let the chord AB cut by circle C be the origin of the circle with the diameter of L, and find out the equation of the straight line L

X & # 178; + Y & # 178; - 2x + 4y-4 = 0 (x - 1) & # 178; + (y + 2) & # 178; = 9C (1, - 2), radius r = 3, let the midpoint of AB be D, the radius of the new circle be RCD, the slope is - 1, the equation is y + 2 = - (x - 1), y = - X - 1, straight line L: y = x + B, X - y + B = 0, intersection D (- (B + 1) / 2, (B - 1) / 2

If the circle C: x2 + (Y-3) 2 = 9 is known and the chord op of circle C is made through the origin, then the trajectory equation of the midpoint Q of OP is ()
A. （x-32）2+y2=94（y≠0）B. （x-32）2+y2=94C. x2+（y-32）2=94（y≠0）D. x2+（y-32）2=94

Let Q (x, y) (Y ≠ 0), then p (2x, 2Y), substituting the circle C: x2 + (Y-3) 2 = 9, we can get 4x2 + (2y-3) 2 = 9. The trajectory equation of Q is x2 + (y-32) 2 = 94 (Y ≠ 0). So the answer is: x2 + (y-32) 2 = 94 (Y ≠ 0)

The line y = MX and the circle x ^ 2 + y ^ 2 + 8x-6y + 21 = 0 intersect at P, Q. find the op vector times the OQ vector

X ^ 2 + y ^ 2 + 8x-6y + 21 = 0 let P (x1, MX1) Q (x2. MX2) vector OP point multiplication OQ = x1x2 + m ^ 2x1x2 = (m ^ 2 + 1) x1x2x1x2 be two solutions of the equation x ^ 2 + m ^ 2x ^ 2 + 8x-6mx + 21 = 0, x1x2 = 21 / (m ^ 2 + 1) vector OP point multiplication OQ = x1x2 + m ^ 2x1x2 = (m ^ 2 + 1) x1x2 = 21

When point a moves on the curve y = x ^ 2 + 3, connect point a and fixed point B (6,0). Find the trajectory equation of the midpoint m of ab
rtrrrrrrrrrrrrrrrrrrrrrrrrr

A(a,b)
So m [(a + 6) / 2, (B + 0) / 2]
So x = (a + 6) / 2, a = 2x-6
y=(b+0)/2,b=2y
Put a in
b=a^2+3
4y^2=(2x-6)^2+3
4y^2=4x^2-24x+33

(1) When a moving point P moves on circle x2 + y2 = 4, find the trajectory equation of the midpoint m of the line connecting point P and fixed point a (4,3). (2) find the trajectory equation of the midpoint n of chord BC from fixed point a (4,3) leading to secant ABC of circle x2 + y2 = 4. (3) in plane rectangular coordinate system xoy, the intersection points of curve y = x2-6x + 1 and coordinate axis are on circle C, B, and OA ⊥ ob, find the value of A

(1) Let the coordinates of midpoint m be (x, y), and the coordinates of point p be (2X-4, 2y-3) obtained from the midpoint coordinate formula. The equation about X and Y obtained by substituting the coordinates of point P into the circle is the trajectory equation of midpoint m (because point P is on the circle), that is, (2X-4) 2 + (2y-3) 2 = 4; (2) let the coordinates of midpoint n be (x, y), and the center of the circle be o, then on ⊥ AC, and the center coordinates are (0, 0), and the Yes From KAC = y − 3x − 4, Kon = YX, because on ⊥ AC, so KAC · Kon = - 1, that is, y − 3x − 4 · YX = - 1, we can get (X-2) 2 + (y-32) 2 = 254; (3) according to the meaning of the Title, we can set the center of the circle as (3, b). From y = x2-6x + 1, let x = 0, then y = 1; from y = 0, then x = 3 ± 22, so, (3-0) 2 + (B-1) 2 = (± 22) 2 + B2, we can get b = 1, then (± 22) 2 + B2 = 9, so the equation of circle C is (x-3) ）2 + (Y-1) 2 = 9. ② let coordinates: a (x1, Y1), B (X2, Y2), a and B satisfy the simultaneous equations of straight line X-Y + a = 0 and circle (x-3) 2 + (Y-1) 2 = 9. Eliminate y and get 2x2 + (2a-8) x + a2-2a + 1 = 0. If there are two intersection points of a and B, that is, two solutions of the equation, then △ = 56-16a-4a2 > 0, so there are X1 + x2 = 4-A, x1x2 = A2 − 2A + 12. ③ from OA ⊥ ob, x1x2 + y1y2 = 0, and Y1 = X1 + A, y2 = x 2 + A, i.e. x1x2 + a (x1 + x2) + A2 = 0. ④ substitute ④ into ③ to get a = - 1, and substitute it into △ = 56-16a-4a2 for test, △ = 56 + 16-4 = 68 ＞ 0, which is consistent. So a = - 1

If a moving point P moves on the circle x ^ 2 y ^ 2 = 1, then the trajectory equation of the midpoint of the line between point P and the fixed point (3,0) is obtained

Let the midpoint be (x, y)
From the midpoint coordinate formula
Then p (2x-3,2y)
P on a known circle
（2x-3）²+（2y）²=1
（x-3/2）²+y²=1/4

When a moving point P moves on the circle x 2 + y 2 = 1, the trajectory equation of the midpoint m connecting it with the fixed point a (3,0) is obtained

Take any point B (m, n) on the circle x2 + y2 = 1, and let the midpoint m (x, y) of line AB have x = 3 + M2y = 0 + N2, that is, M = 2x-3n = 2Y. According to M2 + N2 = 1, we can get (x-32) 2 + y2 = 14, that is, the trajectory equation of midpoint m is (x-32) 2 + y2 = 14

The trajectory equation of the midpoint of the line connecting point P (4, - 2) and any point of circle x2 + y2 = 4 is ()
A. （x-2）2+（y+1）2=1B. （x-2）2+（y+1）2=4C. （x+4）2+（y-2）2=1D. （x+2）2+（y-1）2=1

Let any point on the circle be (x1, Y1) and the midpoint be (x, y), then x = X1 + 42y = Y1 − 22x1 = 2x − 4Y1 = 2Y + 2, substitute x2 + y2 = 4 to get (2X-4) 2 + (2Y + 2) 2 = 4, and simplify to (X-2) 2 + (y + 1) 2 = 1