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

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

A (x1, Y1), B (X2, Y2), P (2, - 3) P is the AB midpoint: 1) X1 + x2 = 2 * 2, namely X1 + x2 = 42) Y1 + y2 = 2 * (- 3) Y1 + y2 = 2 * (- 3) namely Y1 + y2 = - 6a, B on the circle: 3) X1 ^ 2 + Y1 ^ 2 = 204) x2 ^ 2 + Y2 ^ 2 = 203) - 4) (x1-x2) * 4 = (y2-y1) * (- 6) simplify, (y1-y2) / (x1-x2) = 4 / 6 = 2 / 3, this is the slope slope slope slope slope slope slope slope slope slope slope slope slope is the slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope slope 2

If point P is the midpoint of chord AB, then the equation of the line where chord AB is located is______ .

The point P (1,2) is inside the circle C (X-2) 2 + y2 = 9,
∵ point P is the midpoint of the chord ab,
∴CP⊥AB，
The slope of chord AB is k = − 1
KCP=−1
2−0
1−2=1
2，
The equation of the line where the string AB is located is Y-2 = 1
2（x-1），
That is: x-2y + 3 = 0,
So the answer is: x-2y + 3 = 0

If the chord ab of the circle x ^ 2 + y ^ 2 = 20 is made through the point P (2, - 3), and | ab | = 8, then the equation of the line where the chord AB is located is?

Let the equation of the line passing through the point p be y + 3 = K (X-2), and its intersection point with the circle is (x1, Y1), (X2, Y2) by introducing y = kx-2k-3 into x? + y? = 20, we obtain: (1 + k? X 2) - (4K? + 6K) x + 4K? + 12k-11 = 0x1 + x2 = (4K? + 6K) / (1 + k?), X1

Given that P (2, - 1) is the midpoint of chord ab of circle (x-1) ^ 2 + y ^ 2, then the equation of the line where chord AB is located is?

The coordinates of the center of the circle (x-1) ^ 2 + y ^ 2 = R ^ 2 are (1,0), set to o,
Then the slope of OP K1 = (2-1) / (- 1) = - 1
OP is perpendicular to AB, so the slope of AB is K2 = - 1 * 1 / (- 1) = 1
So the linear equation of AB is:
y+1=x-2
=》y=x-3

The straight line L intersects the circle x ^ 2 + y ^ 2 + 2x-4y + 1 = 0 at the two points of ab. if the midpoint of chord AB (- 2,3), then the equation of line L is X-Y + 5 = 0

x^2+y^2+2x-4y+1=0
x^2+2x+1+y^2-4y+4=4
(x+1)^2+(y-2)^2=4
Center C (- 1,2),
Set point d (- 2,3)
From the vertical diameter theorem,
L is perpendicular to CD
k(CD)=(3-2)/(-2+1)=-1
The slope of L is 1
∴y-3=1*(x+2)
x+2-y+3=0
x-y+5=0

The intersecting chord equation of two circles x2 + Y2 + 4x-4y = 0 and X2 + Y2 + 2x-12 = 0 is A。 x+2y-6=0 B.x-3y+5=0 C.x-2y+6=0 D.x+3y-8=0

Subtract two circles
Answer c

Given that the two circles x2 + Y2 + 4x-4y-1 = 0 and X2 + Y2 + 2x + 2y-2 = 0 intersect at P and Q, then the equation of the perpendicular line of the common chord PQ is

The problem is equivalent to the equation of a straight line passing through the center of two circles
The two circle equations are reduced to (x + 2) 2 + (Y-2) 2 = 9 and (x-1) 2 + (Y-1) 2 = 4
So the centers of the two circles are (- 2,2), (1,1)
The linear equation obtained is x + 3y-4 = 0
Good afternoon. I don't know. Please ask!

In the line passing through point (2,1), the equation of the line with the longest chord cut by the circle x2 + y2-2x + 4Y = 0 is X2: the square of X

(x-1)^2+(y+2)^2=5
Center (1, - 2)
The longest string is the diameter
Therefore, the straight line passing through (2,1) and the center of the circle is the required line
(y-1)/(-2-1)=(x-2)/(1-2)
3x-y-5=0

If the straight line kx-y + 2 = 0 and the circle x ^ 2 + y ^ 2 = 1 intersect at two points a and B, then the locus of the midpoint of chord AB is

Method 1, let the midpoint of AB be p (x, y), a (x1, Y1), B (X2, Y2)
From kx-y + 2 = 0, y = KX + 2
Substituting it into the equation of a circle
x^2+(kx+2)^2=1
(k^2+1)x^2+4kx+3=0
Then X1 + x2 = - 4K / (k ^ 2 + 1)
The coordinates of point P x = (x1 + x2) / 2 = - 2K / (k ^ 2 + 1) (1)
Because the point P is on the line AB: y = KX + 2
Then k = (Y-2) / X II.
Replace (2) with (1) and sort out
x^2+(y-1)^2=1
That is to say, the locus of point P is a circle with radius 1 (the part within the known circle) with (0,1) as its center
Method 2. The oblique section of the line kx-y + 2 = 0 is y = KX + 2
Then compared with the y-axis, the vertex m (0,2)
Let the midpoint of AB be p and connect Op,
Then op ⊥ AB,
So △ MPO is a right triangle
Therefore, the locus of point P is the part of the circle whose diameter is mo within the known circle
The center of the circle is the midpoint of OM (0,1), and the radius is OM / 2 = 1
The trajectory equation is x ^ 2 + (Y-1) ^ 2 = 1 (0)
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Let the locus of the midpoint of the chord cut by the line kx-y + 1 = 0 by a circle: x ^ 2 + y ^ 2 = 4 is C, then the position relationship between the curve C and the straight line x + Y-1 = 0 is obtained

intersect