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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
(1) ∵ KAB = 1, the coordinates of the midpoint of AB are (1,2) ᙽ the equation of the straight line CD is: Y-2 = - (x-1), i.e. x + Y-3 = 0; (2) if the center of the circle P (a, b), then a + B-3 = 0 is obtained from P on CD. ① and the diameter | CD | = 410, | PA | = 210 ﹤ (a + 1) 2 + B2 = 40; ② ① substituting ② to eliminate a, we can get b2-4b-12 = 0, and the solution is b =
It is known that the center of the circle P is in the second quadrant and passes through a (- 1,0) and B (3,4). The vertical bisector of line AB intersects the circle P at points c and D, and CD = 4 pieces of 10. (1) find the equation of circle P. (2) suppose the point q is on the original P, how many points are there for △ qAB to get Mina and point Q equal to 8? Prove your conclusion
(1) The center of the circle is on the straight line CD, so the radius is two 10,
The coordinates of intersection point of line AB and CD are (3-1) / 2 = 1, (4 + 0) / 2 = 2, that is, the coordinates of intersection point are (1,2),
And the linear equation of AB is X-Y + 1 = 0, so the slope of line CD is - 1. The equation of straight line CD is x + Y-3 = 0 from the point oblique formula. Therefore, we can set the coordinates of the center of the circle as (x, - x + 3). A right triangle is formed by the distance from the center of the circle to the line AB and the half and radius of the chord ab
So the equation of circle P is (x + 3) ^ 2 + (y-6) ^ 2 = 40
(2) The linear equation of AB is X-Y + 1 = 0, the length of AB is 4 pieces 2, let Q (x0, Y0), the distance from point Q to line AB is h,
Because the area of △ qAB is 8, so four 2 * 0.5 * H = 8, the solution H = 2 2 2,
So the distance from point Q to line AB is x0-y0 + 1 / root 2 = 2,
The result shows that x0-y0 = 3 or x0-y0 = - 5
There is a point P (- 1,2) in the circle (x + 1) ^ 2 + y ^ 2 = 8, and ab passes through the point P (1). If the absolute value of chord length AB = 2, the root sign 7, find the inclination angle a of the straight line ab (2) If there are exactly three points on the circle and the distance from the line AB is equal to the root 2, find the equation of the line ab
(x+1)²+y²=8
Center (- 1,0)
Let AB lie on the line y = K (x + 1) + 2
Chord center distance = root (8-7) = 1
Chord center distance = distance from center of circle to straight line y = K (x + 1) + 2
Under | 2 | radical sign (K | + 1) = 1
K = ± root 3
2 radius = 2 under 2 roots
Chord center distance = 2 under radical
So | 2 | / under root sign (K | 1) = 2 under root sign
k=±1
So the line is y = x + 3 or y = - x + 1
It is known that the point centered on point m passes through points a (- 1,1) and B (3,5), and the vertical bisector of line AB intersects the circle m at two points c and D, and CD = 2 times the root sign 10
If CD is the diameter, then the radius is √ 10. If the circle M: (x + a) + (y + b) = 10, then: (A-1) + (B + 1) = 10 (1) (a + 3) + (B + 5) = 10 (2), then 4 (2a + 2) = - 4 (2B + 6) a + 1 = B + 3 a = B + 2 (*) replace (*) with (1) 2 (B + 1) = 10, B + 1 = ± √ 5, B = ±√ 5-1
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 and changes sign 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 are there that make the area of the triangle qAB equal to 8? Prove your conclusion
1) Because the straight line CD is the vertical bisector of line AB, and the slope of line AB is 1, the slope of line CD is - 1,
The midpoint of line AB is on the line CD, and the coordinates of the midpoint are (1,2), so the equation of line CD is x + Y-3 = 0
2) The center of the circle is on the straight line CD, so the radius is two 10,
The coordinates of intersection point of line AB and CD are (3-1) / 2 = 1, (4 + 0) / 2 = 2, that is, the coordinates of intersection point are (1,2),
And the linear equation of AB is X-Y + 1 = 0, so the slope of line CD is - 1. The equation of straight line CD is x + Y-3 = 0 from the point oblique formula. Therefore, we can set the coordinates of the center of the circle as (x, - x + 3). A right triangle is formed by the distance from the center of the circle to the line AB and the half and radius of the chord ab
So the equation of circle P is (x + 3) ^ 2 + (y-6) ^ 2 = 40
(3) The linear equation of AB is X-Y + 1 = 0, the length of AB is 4 pieces 2, let Q (x0, Y0), the distance from point Q to line AB is h,
Because the area of △ qAB is 8, so four 2 * 0.5 * H = 8, the solution H = 2 2 2,
So the distance from point Q to line AB is x0-y0 + 1 / root 2 = 2,
The result shows that x0-y0 = 3 or x0-y0 = - 5
As shown in the figure, the radius od of circle O is perpendicular to the chord AB at point C, connecting AO and extending the intersection circle O at point E, connecting EC. If AB = 8 and CD = 2, how long is EC
The center of the circle is o
Connect be
Because the radius od of circle O is perpendicular to the chord AB at point C, ab = 8
So AC = BC = 4 OD, vertical AC
Let the radius od be x (AO = x)
Because CD = 2
In triangular AOC, the square of AC + the square of OC = the square of Ao
So the square of AC + (od-cd) is the square of Ao
So the square of 4 + the square of (X-2) equals the square of X
X = 5
Diameter AE = 10
Because we connect AO and extend the intersection circle O at point E
So the angle Abe is 90 degrees
So in the triangle Abe, EB = 6
Because BC = 4
So in the triangle CBE, EC = root 52
As shown in the figure, in circle O, the length of chord AB is 8, OD is perpendicular to AB, intersection AB is at point D, intersection circle O is at point C, OD ratio CD = 1:2, find the length of CD
As shown in the figure AOod:cd=1 : 2 then od: OC = 1:3ao = Co ᙽ OA: od = 3:1
In circle C, can we get the value of vector ab · vector BC only by knowing the radius of circle C or the length of chord AB
AB*BC=-0.5*|AB|^2
It is proved that in circle C, we only need to know the radius of circle C or the length of chord ab
Just the length of the string ab
Let the angle between vector AB and vector BC be B
Then vector ab · vector BC = - | ab | BC | CoSb = - | ab | BC | (| ab | / 2) / | BC | = - | ab | / 2
If there is an ab chord with chord length of 6 in the circle with radius 6, then the length of the arc to which the chord AB is opposite is etc.
1/2 * 1/3π *R=π
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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)
- 4. 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
- 5. 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
- 6. 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
- 7. It is known that AB is the diameter of circle O, PD tangent circle O to C, and the extension line of Ba intersects PC at P Angle P = 26 degrees, calculate angle BCD
- 8. As shown in the figure of triangle ABC, ad is the center line on BC side, f is a point on ad, and AF: FD = 1:3, extend BF, cross AC to e, and find AE: EC
- 9. As shown in the figure, ad is the diameter of circumscribed circle of △ ABC, CE ⊥ ad intersects ad with F and ab with E. try to explain that AC square = ab × AE
- 10. As shown in the figure, AB is the diameter of semicircle o, and point C is a point on the semicircle. Through point C, make CD perpendicular to D, AC = 2 times, root sign 3cm, ad: DB = 3:1, and find the length of AD As shown in the figure, AB is the diameter of semicircle o, and point C is a point on the semicircle. Through point C, make CD perpendicular to D, AC = 2 times, root sign 3cm, ad: DB = 3:1, and find the length of AD
- 11. 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
- 12. 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
- 13. 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?
- 14. 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
- 15. 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
- 16. 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
- 17. If x ^ 2 + y ^ 2-2y = 0, the midpoint of chord AB is (- 1 / 2,3 / 2), then ab=
- 18. 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
- 19. 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
- 20. 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