If P9 (2, - 1) is the midpoint of the chord ab of the circle (x-1) 2 + y2 = 25, what is the equation of AB

If P9 (2, - 1) is the midpoint of the chord ab of the circle (x-1) 2 + y2 = 25, what is the equation of AB

From the equation of the circle, the coordinates of the center C of the circle can be obtained as (1,0)
Obviously, the line CP is perpendicular to the line ab
The slope of straight line CP is k = (- 1-0) / 2-1 = - 1
So the slope of the line AB is k = 1
The equation of AB is y + 1 = X-2
That is, x-y-3 = 0

If P (2, - 1) is the midpoint of chord ab of circle (x-1) 2 y2 = 25, then the equation of line AB is given

Solution: a (x1, Y1), B (X2, Y2), then: X1 + x2 = 4, Y1 + y2 = -2, (x1-1) + Y1 = 25 (1), (x2-1) + y2 = 25 (2), (1) - - (2) get, (x1-1 + x2-1) (x1-1-x2 + 1) + (y1-y2) (y1-y2) (Y1 + Y2) (Y1 + Y2) (Y1 + Y2) (Y1 + Y2) = 0,2 (x1-x2) + 2 (y1-y2) = 0, (y2-y1) / (x2-x1) / (x2-x1) = - 1 = k, (y + y + 1 = (X-2), (x + y-y-y-2), (x + y-y-y-y-2), (X-2) (x-1 the equation of AB is: x + Y-1 = 0

If P (2, - 1) is the midpoint of chord ab of circle (x-1) 2 + y2 = 25, then the equation of line AB is () A. x-y-3=0 B. 2x+y-3=0 C. x+y-1=0 D. 2x-y-5=0

The center of the circle is known to be o (1, 0)
Kop = 0 + 1
1−2＝−1
kABkOP=-1
K AB = 1, and the straight line AB crosses the point P (2, - 1),
The equation of line AB is x-y-3 = 0
So choose a

Let the equation of circle be x2 + y2-4x-5 = 0, (1) Find the center coordinates and radius of the circle; (2) If the midpoint of a chord ab of this circle is p (3,1), find the equation of line ab

(1) The formula of x2 + y2-4x-5 = 0 was as follows: (X-2) 2 + y2 = 9
The coordinate of the center of the circle is C (2.0), and the half meridian is r = 3 (6 points)
(2) Let the slope of line AB be K
According to the knowledge of circle, CP ⊥ AB,  KCP · k = - 1
KCP = 1 − 0
3−2=1，∴k=-1．
The equation of line AB is Y-1 = - 1 (x-3)
That is: x + y-4 = 0 (12 points)

It is known that there is a point P (- 2,1) in the circle x square + y square = 8, AB is the chord (1) passing through the point P and the inclination angle is a (1). When a = 135 °, find the equation of ab (2) If the chord AB is bisected by point P, find the equation of line ab

Is p (- 1,2)?
(1) When α = 135 °, k = - 1, and then pass through point P (- 2,1)
Then Y-1 = - 1 (x + 2)
Y + X + 1 = 0
X squared + y squared = 8
X1, Y1 and X2 Y2 are obtained
Then AB = √ (x1-x2) square + (y1-y2) square

There is a point P (- 1,2) in the circle (x + 1) ^ 2 + y ^ 2 = 8, AB passes through the point P, (1) find the equation that the chord AB is the shortest line (2) If there are exactly three points on the circle and the distance from AB is equal to the root two, find the equation of straight line ab

hello
(1) If the center of a circle is C, then C (- 1,0)
Chord AB over point P
When CP is perpendicular to AB, the chord AB is the shortest
From the coordinates, CP is perpendicular to the x-axis
So AB is perpendicular to the Y axis
So the equation of AB is y = 2
(2) If there are exactly three points on the circle, the distance from AB is equal to root two
Then the distance from the intersection point of the perpendicular line of the straight line AB and the circle to AB is √ 2
Let AB focus on D
Connect CD
According to the meaning of the title, CD = 2 √ 2 - √ 2 = √ 2
Because the slope of the line must exist
Therefore, let the equation of the straight line be Y-2 = K (x + 1)
That is, kx-y + K + 2 = 0
C (- 1,0) according to the distance formula from point to line,
CD=|-k+k+2|/√(k²+1)=√2
K = 1 or K = - 1
So AB's equation is X-Y + 3 = 0
Or x + Y-1 = 0

There is a point P (- 1,2) in the circle x square + y square = R square. AB is the chord passing through the point P. when AB is the shortest time, find the equation of the straight line ab

Ideas:
The shortest time AB is perpendicular to Op
The slope of OP is - 2, so AB has a slope of 1 / 2
answer:
Y = 1 / 2 * x + 5 / 2 (or 2y-x = 5)

If the chord ab of circle x * 2 + y * 2 = 20 is made through point P (2, - 3), and P is bisected into AB, then the equation of the line where chord AB is located is

A(x1,y1),B(x2,y2),P(2,-3)
P is the midpoint of ab
1) X1 + x2 = 2 * 2, that is, X1 + x2 = 4
2) Y1 + y2 = 2 * (- 3), that is, Y1 + y2 = - 6
A. B on the circle:
3)x1^2+y1^2 = 20
4)x2^2+y2^2 = 20
3) - 4)
(x1-x2)*4 =(y2-y1) * (-6)
Simplification,
(y1-y2)/(x1-x2) = 4/6 = 2/3
This is the slope of 2 / 3, point P (2, - 3), using the point oblique form to write the line equation
y+3 = 2/3 * (x-2)

If the chord ab of the circle x ^ 2, y ^ 2 = 20 is made through the point P (2, - 3), and P bisects AB, then the equation of the line where the chord AB is located is

The answer is 2x-3y-13 = 0
You can first find AB's perpendicular line, that is, the slope of OP is - 2 / 3, then find AB slope, and then substitute P

The equation of the known circle is: x ^ 2 + y ^ 2 = 4 (1) the straight line L passes through the point (1,2), and intersects with the circle at two points a and B, if AB = 2, root sign 3

Let the linear l equation be: x + ay + B = 0, the line L passes through the point (1,2): 1 + 2A + B = 0... (1) the radius of the circle is r = 2, the center of the circle is at the origin o (0,0) AB = 2 = = > the distance from the center O to the line L = √ [R ^ 2 - (AB / 2) ^ 2] = 1, that is, | 0 + 2 * 0 + B | / √ (1 ^ 2 + A ^ 2) = 1... (2) (1) (2) = > (a, b) = (0, - 1), (-...)