Given that the line L: 2x-y-1 = 0 and the circle C: x2 + y2-2y-1 = 0 intersect at two points AB, then the chord length ab=

Given that the line L: 2x-y-1 = 0 and the circle C: x2 + y2-2y-1 = 0 intersect at two points AB, then the chord length ab=

x²+(y-1)²=2
r=√2
Center (0,1)
Then the chord center distance d = | 0-1-1 | / √ (2 | 1 |) = 2 / √ 5
From Pythagorean theorem
Chord length = 2 √ (r? - D?) = 2 √ 30 / 5

Given the circle C: the square of X + the square of Y - 4x + 2Y + 1 = 0, on the symmetric circle D of the straight line L: x-2y + 1 = 0, find the equation of circle D

X ^ 2 + (Y-3) ^ 2 = 4 is enough to find the center of symmetry. From symmetry, K1 * K2 = - 1 is obtained. Let y = - 2x + A. substitute (2, - 1) into a to find the center of the circle

Given that the square of the circle x + the square of Y - 2x + 2y-3 = 0 and the circle x square + y square + 4x-1 = 0 are symmetric about the straight line L, we can find the equation of the straight line L

According to the meaning of the title, the centers of the two circles are
A（1,-1）,B（-2,0）
Then the slope of line AB is k = (- 1-0) / (1 - (- 2)) = - 1 / 3
And the coordinates of the midpoint of a and B (- 1 / 2, - 1 / 2)
So the slope of the line L is 3, and it passes through the point (- 1 / 2, - 1 / 2)
Then the equation of the line L is y + 1 / 2 = 3 (x + 1 / 2)
That is, 3x-y + 1 = 0

The straight line passing through the point P (- 2,3) is cut by the circle x ^ 2 + y ^ 2-4x + 2y-2 = 0. Find the equation of the line where the longest chord is located As the title

The longest chord is the diameter of the circle
x^2+y^2-4x+2y-2=0
（x-2）^2+(y+1)^2=7
Center (2, - 1)
Slope of line = - 1
Linear equation
y-3=-(x+2)
y=-x+1

Given that [(square of X + square of Y) - (square of X-Y) + 2Y (x + y)] / 4Y = 1, find the value of 4x / 4x squared - y squared - 1 / 2x + y

Original formula = 4
(because the equation is quadruple)

Simplification evaluation: 4x square - 8xy + 4Y square / 2x square - 2Y square, where x = 2, y = 3

The original formula = 4 (X-Y) square / 2 (x + y) (X-Y) = 2 (X-Y) / x + y
=-2/5

Given that x2-4x + 2y2-4y + 6 = 0, simplify and then evaluate: (2x + y) (2x-y) - (2x-y) 2

x2-4x+2y2-4y+6=x2-4x+2y2-4y+2+4=（x-2）2+2（y-1）2=0，
∴x-2=0，y-1=0，
X = 2, y = 1,
∵（2x+y）（2x-y）-（2x-y）2，
=4x2-y2-（4x2-4xy+y2），
=4x2-y2-4x2+4xy-y2，
=-2y2+4xy，
=-2+8=6．

Given that x2-4x + 2y2-4y + 6 = 0, simplify and then evaluate: (2x + y) (2x-y) - (2x-y) 2

x2-4x+2y2-4y+6=x2-4x+2y2-4y+2+4=（x-2）2+2（y-1）2=0，
∴x-2=0，y-1=0，
X = 2, y = 1,
∵（2x+y）（2x-y）-（2x-y）2，
=4x2-y2-（4x2-4xy+y2），
=4x2-y2-4x2+4xy-y2，
=-2y2+4xy，
=-2+8=6．

Simplify and evaluate the square of 8x - 5x (4y-x) + 4x × (- 4x + 5 / 2Y) x = 1, y = 3

8x^2-20xy+5x^2-16x^2+10xy
-3x^2-10xy
x=1,y=3
-3-30=-33

Simplify and then evaluate 4x ^ 2-8xy + 4Y ^ 2 / 2x ^ 2-2y ^ 2, where x = 2, y = 3

(4x^2-8xy+4y^2)/(2x^2-2y^2)
=[4(x^2-2xy+y^2)]/[2(x^2-y^2)]
=[4(x-y)^2)]/[2(x-y)(x+y)]
=2(x-y)/(x+y)
=2*(2-3)/(2+3)
=-2/5