If x ^ 2 + y ^ 2-2y = 0, the midpoint of chord AB is (- 1 / 2,3 / 2), then ab=

x^2+y^2-2y=0
x^2+(y-1)^2=1
The distance from the center of a circle to the midpoint of AB is d = √ [(0 + 1 / 2) ^ 2 + (1-3 / 2) ^ 2] = √ 2 / 2
AB=2√(r^2-d^2)=2√(1-1/2)=√2

If the circle center P is on the straight line y = x and tangent to the line x2y-1 = 0, the chord AB length obtained by the circle section of the upper half axis of the Y axis is 2

If the center of the circle is on y = x, let the equation of the circle be (x-a) ^ + (Y-A) ^ = R ^ (sign ^ denotes the square) and bring in x = 0, then a ^ + (Y-A) ^ = R ^. The chord AB length of Y ^ - 2ay + 2A ^ - R ^ = 0 is 2, and (y1-y2) ^ = 4. According to Weida's theorem, Y1 + y2 = 2A, Y1 * y2 = 2A ^ - R ^ (y1-y2) ^

The straight line L passes through the point P (5,5) and intersects the circle C: X squared + y squared = 25. The chord length is 4 and the sign is 5. Find the equation of L? Please thank you

Drawing shows that the distance between the line L and the center of the circle is √ 5. Let the equation of the straight line be Y-5 = K (X-5). According to the formula of the distance from the center of the circle to the straight line, we can get k = 1 / 2, or K = 2, so the equation of the line L is y = x / 2 + 5 / 2 or y = 2x-5

If the straight line L passes through the point (- 5,10) and cuts the chord length on the circle x ^ 2 + y ^ 2 = 25, then what is the equation of the straight line l Hope to have an analysis

According to the point oblique type
Let the linear equation be Y-10 = K (x + 5)
That is, y = KX + 5K + 10
Replace the circle equation x ^ 2 + y ^ 2 = 25
X ^ 2 + (KX + 5K + 10) ^ 2 = 25
Expansion (1 + k) x ^ 2 + (10K ^ 2 + 20k) x + (25K ^ 2 + 100k + 75) = 0
According to the chord length formula
L=√(1+k^2)√[(x1+x2)^2-4x1x2]
Then K can be obtained

Given that the straight line L crosses the point (- 5, - 10) and cuts the chord length on the circle x ^ 2 + y ^ 2 = 25, the chord length is 5 root sign 2, and the equation of the straight line L is solved

Let the equation of the line l be: y + 10 = K (x + 5)
That is, kx-y + 5k-10 = 0
Because the center of the circle x ^ 2 + y ^ 2 = 25 is (0,0) and the radius is r = 5
Let the distance between the center of the circle and the line l be D, and the chord length L = 5 √ 2
Then D 2 = R 2 - (L / 2) 2
So D? 2 = 5? 2 - (5 √ 2 / 2) 2
That is, d = 5 √ 2 / 2
From the distance formula of circle center to straight line L
d=|5k-10|/√(k²+1)=5√2/2
The solution is: k = 1 or K = 7
So the equation of the straight line L is: y + 10 = (x + 5) or y + 10 = 7 (x + 5)
That is, y = X-5 or y = 7x + 25

Given that the radius of the circle is root 10, the center of the circle is on the straight line y = 2x and cut by the line X-Y = 0, the chord length is 4 times the root sign 2, and the equation of the circle is solved

Let the coordinates of the circle center be (x, 2x), and the distance from the center of the circle to X-Y = 0 is d. because the length of the truncated chord is 4 √ 2 and the radius of the circle is √ 10, from the Pythagorean theorem there is (2 √ 2) ^ 2 + D ^ 2 = (√ 10) ^ 2, then d ^ 2 = 10-8 = 2, and the formula of distance from point to line has d = | x-2x | / √ [1 ^ 2 + (- 1) ^ 2] = | x | / √ 2, so there is x ^ 2 / 2 = 2, x = ± 2

The radius of the circle is known to be The center of the circle is on the line y = 2x, and the chord length of the circle cut by the line X-Y = 0 is 4 2. Find the equation of circle

Let the center of the circle (a, 2a) be set, and the chord center distance D can be obtained from the chord length formula=
10−8=
2，
From the distance formula of point to line, d = | a − 2A|
2=
Two
2|a|，
The center coordinate of the circle is (2,4), or (- 2, - 4), and the radius is
10，
The equation of the circle is: (X-2) 2 + (y-4) 2 = 10 or (x + 2) 2 + (y + 4) 2 = 10

The radius of the circle is known to be The center of the circle is on the line y = 2x, and the chord length of the circle cut by the line X-Y = 0 is 4 2. Find the equation of circle

Given that the circle C passes through the point (1, 0), and the center of the circle is on the positive half axis of the X axis, the chord length of the straight line L: y = X-1 cut by the circle is 2 2, then the standard equation of circle C is () A. （x+1）2+y2=4 B. （x-3）2+y2=4 C. （x-1）2+y2=4 D. （x+3）2+y2=4

Let the coordinates of the center of the circle C be (a, 0), a > 0, then the distance from the center of the circle to the straight line L: y = X-1 is d = | a − 0 − 1|
2=|a−1|
2．
Because the radius r = | A-1|=
(a−1
2)2+(
2) 2, a = 3, or a = - 1 (omitted),
Therefore, the center of circle C is (3,0), and the radius is 3-1 = 2. Therefore, the standard equation of circle C is (x-3) 2 + y2 = 4,
Therefore, B

If x ^ 2 + y ^ 2-2y = 0, the midpoint of chord AB is (- 1 / 2,3 / 2), then ab=