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Let the midpoint of chord ab of circle x2 + y2-4x-5 = 0 be p (3,1), then the equation of straight line AB is () A. x+y-4=0 B. x+y-5=0 C. x-y+4=0 D. x-y+5=0
The center O (2,0) of the circle x2 + y2-4x-5 = 0,
The midpoint of chord ab of circle x2 + y2-4x-5 = 0 is p (3,1),
The line AB is perpendicular to the line Op,
∵kOP=1−0
3−2=1，∴kAB=-1，
The equation of line AB is: Y-1 = - (x-3), and the result is: x + y-4 = 0
Therefore, a
If point P (3,1) is the midpoint of chord ab of circle x2 + y2-4x-21 = 0, then the equation of line AB is______ .
By transforming the equation of a circle into a standard equation, we get: (X-2) 2 + y2 = 25,
So the o coordinate of the center of the circle is (2,0), and the radius of the circle is r = 5,
According to the vertical diameter theorem, Op ⊥ AB, P (3,1), ﹥ Kop = 1 − 0
3−2=1，
Then K AB = - 1, and the straight line AB passes through point P,
So the equation of line AB is: Y-1 = - 1 (x-3), that is, x + y-4 = 0
So the answer is: x + y-4 = 0
Let the midpoint of chord ab of circle x2 + y2-4x-5 = 0 be p (3,1), then the equation of straight line AB is () A. x+y-4=0 B. x+y-5=0 C. x-y+4=0 D. x-y+5=0
The center O (2,0) of the circle x2 + y2-4x-5 = 0,
The midpoint of chord ab of circle x2 + y2-4x-5 = 0 is p (3,1),
The line AB is perpendicular to the line Op,
∵kOP=1−0
3−2=1，∴kAB=-1，
The equation of line AB is: Y-1 = - (x-3), and the result is: x + y-4 = 0
Therefore, a
Let the midpoint of chord ab of circle x2 + y2-4x-5 = 0 be p (3,1), then the equation of straight line AB is () A. x+y-4=0 B. x+y-5=0 C. x-y+4=0 D. x-y+5=0
The center O (2,0) of the circle x2 + y2-4x-5 = 0,
The midpoint of chord ab of circle x2 + y2-4x-5 = 0 is p (3,1),
The line AB is perpendicular to the line Op,
∵kOP=1−0
3−2=1，∴kAB=-1，
The equation of line AB is: Y-1 = - (x-3), and the result is: x + y-4 = 0
Therefore, a
Let the midpoint of chord ab of circle x2 + y2-4x-5 = 0 be p (3,1), then the equation of straight line AB is () A. x+y-4=0 B. x+y-5=0 C. x-y+4=0 D. x-y+5=0
The center O (2,0) of the circle x2 + y2-4x-5 = 0,
The midpoint of chord ab of circle x2 + y2-4x-5 = 0 is p (3,1),
The line AB is perpendicular to the line Op,
∵kOP=1−0
3−2=1，∴kAB=-1，
The equation of line AB is: Y-1 = - (x-3), and the result is: x + y-4 = 0
Therefore, a
And the length of the circle passing through the point x = 2 is obtained
First of all, x = 2 is a solution, because point P is on the circle, so the chord length is exactly twice the ordinate of point P. secondly, you can see that there is a line satisfying the condition, which is symmetric with the line x = 2 about the line y = 2x
Through the point P (2, - 3), make the chord ab of the circle x? 2XY? 2 = 24, so that the chord AB is bisected by the point P, then the equation of the line where the chord AB is located is
If point P is in a circle, then the straight line where the chord passing through point P and bisected by point P is perpendicular to the line connecting the center of the circle and B, and the slope of the line connecting the center of the circle and B is - 1, then the slope of the straight line is 1, and passing through the point P (2, - 3), the equation of the straight line is: x-y-5 = 0
It is known that the midpoint of the chord ab of the ellipse x? 2 / 36 + y? 2 / 9 = 1 is m (3,1)
Let a (x, y), then B (6-x, 2-y)
A. B on the ellipse, substituting the elliptic equation, we get
x^2/36+y^2/9=1
(6-x)^2/36+(2-y)^2/9=1
The equation of AB is obtained by subtracting the two formulas
(36-12x)/36+(4-4y)/9=0
Namely
36-12x+16-16y=0
3x+4y-13=0
If the chord ab of the circle x ^ 2 + y ^ 2 = 20 is made through the point P (2, - 3), and the chord AB is bisected by the point P, then the equation of the line where AB is located is
P bisecting chord AB is op vertical AB (o is the center of the circle)
kOP=-2/3
So Ka B = 3 / 2
y+3=3/2(x-2)
AB :3x-2y-12=0
If a line passing through the origin intersects with the square of circle x plus the square of Y minus 2x minus 4Y plus 4 equals 0, and the length of the chord obtained is 2, then the equation of the line is
The equation of the circle shows that: (x-1) 2 + (Y-2) 2 = 1
The radius of circle = 1 and the chord length of 2 indicates that the chord is equal to the diameter
The straight line goes through the center of the circle (1,2) and the origin (0,0)
So the line y = 2x
RELATED INFORMATIONS
- 1. What is the chord length of a line passing through the origin and with an inclination angle of 60 degrees cut by the circle x square + y square - 4Y = o?
- 2. 2 ax by + 2 = 0 (a > 0, b > 0). The chord length cut by the circle x ^ 2 + y ^ 2 + 2x + - 4Y + 1 = 0 is 4
- 3. It is known that the length of the two chords of a circle AB CD is two of the equation x ^ 2-42x + 432 = 0, and ab is parallel to CD, and the distance between the two chords is 3. Find the radius
- 4. It is known that the circle centered on point P passes through points a (- 1,0) and B (3,4), and the vertical bisector of line AB intersects the circle P at points c and D, and | CD | = 4 10． (1) Find the equation of straight line CD; (2) Find the equation of circle P; (3) Set the point Q on the circle P. how many points Q have the area of △ qAB equal to 8? Prove your conclusion
- 5. As shown in the figure, the radius of ⊙ o is 1cm, and the lengths of strings AB and CD are respectively 2 cm, 1 cm, then the acute angle α between the strings AC and BD=______ Degree
- 6. As shown in Figure AC is the diameter of circle O, PA is perpendicular to AC, connecting OP, chord CB / / op, diameter BC intersects straight line AC at D, BD = 2PA, prove that BP is tangent of circle O, OP is related to BC Verify the value of sin angle OPA I 1 and can not insert the picture The diameter BC in the problem is changed to the straight line Pb
- 7. As shown in the figure, the vertical bisector of chord AB intersects at point C and intersects chord AB at point D. It is known that ab = 24cm, CD = 8cm (1) Find the circle where the fragment is located (keep the drawing trace and write the method); (2) Find the radius of the circle in (1)
- 8. O is the diameter of a perpendicular point on each other (1) Verification: AC bisection ∠ DAB; (2) If CD = 4, ad = 8, try to find the radius of ⊙ o
- 9. As shown in the figure, the triangle ABC is inscribed with the circle O, ad bisects ∠ BAC, intersects ⊙ O and D, passes through D as de ∥ BC, and intersects the extension line of AC with E The triangle ABC is inscribed with the circle O, ad bisects ⊙ BAC, intersects ⊙ O and D, passes D as de ∥ BC, and intersects the extension line of AC with E Try to judge the position relationship between de and circle O and prove your conclusion 2. If ∠ e = 60 degrees, the radius of circle O is 4, and find the length of ab
- 10. As shown in the figure, it is known that ⊙ o is the circumscribed circle of ⊙ ABC, CE is the diameter of ⊥ o, CD ⊥ AB and D are perpendicular feet. It is proved that ⊙ ACD = ∠ BCE
- 11. Given that the circle C: x 2 + y 2 = 5, the straight line L: ax-y-2a = 0 and circle C intersect at two points AB, the trajectory equation of the midpoint of chord AB is obtained
- 12. It is known that the circle C passes through point 10 and the center of the circle is on the positive half axis of the X axis. The straight line x + Y-1 = 0 is cut by the circle C, and the chord length is 2, and the circle standard is found equation
- 13. If x ^ 2 + y ^ 2-2y = 0, the midpoint of chord AB is (- 1 / 2,3 / 2), then ab=
- 14. Find the equation of the circle whose center is on the line 3x-y = 0, tangent to the x-axis, and cut by the line X-Y = 0
- 15. The straight line kx-y + 1 = 0 and the circle x? 2 + y? 2 = 1 intersect a and B, and find the locus of the midpoint of chord ab
- 16. 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
- 17. Find a point in the ellipse C: x ^ 2 + 4Y ^ 2 = 16. Find the equation of the line where the chord is located with a point a (1, - 1) in the ellipse x * 2 + 4Y * 2 = 16 as the midpoint
- 18. If two circles x2 + y2-10x-10y = 0 and X2 + y2-6x + 2y-40 = 0 intersect at two points, the equation of the line where their common chord is located is______ .
- 19. The linear equation of chord length 8 passing through point a (- 1,10) and cut by circle x2 + y2-4x-2y-20 = 0 is______ .
- 20. Given the circle C1: x square + y square + 2x + 6y + 9 = 0 and circle C2: x square + y square - 6x + 2Y + 1 = 0, find the common tangent equation of circle C1 and circle C2