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Chord length obtained by cutting straight line x-y-5 = 0
It is (y + 5) ^ 2 + y ^ 2-4 (y + 5) + 4Y + 6 = 0y ^ 2 + 10Y + 25 + y ^ 2-4y-20 + 4Y + 6 = 02y ^ 2 + 10Y + 25 + y ^ 2-4y-20 + 4Y + 6 = 02y ^ 2 + 10Y + 11 = 0y1 + 11 = 0y1 + y2 = 10 / 2 = 5y1y2y2 = 11 / 2 = 5y1y2y2 = 11 / 2 (y1-y2) ^ 2 = (Y1 + Y2) ^ 2-4y1y1y2 = 5 ^ 2-4 * 4 * 11 / 2 = 25-22 = 3 (x1-x2) ^ 2 = (Y1 + 5 + 5) (x1-x2) ^ 2 = (Y1 + 5 + 5 + 5) ^ 2 = (y1- y1-5) ^ 2 = (y1-y2) ^ 2 = 3 ℅ chord length = √ [((...)
If (x) - 2 is the length of (2) - 3, then it intersects the root of the line (2)
Note that the chord length of the y-axis intersecting the circle is 2 * √ 3, so x = 0 is one of the lines
The other is a straight line perpendicular to y axis and passing through point a, i.e., y = 3
To sum up, there are x = 0 and y = 3
Pass through point a (11,2) to make the chord of circle x ^ 2 + y ^ 2 + 2x-4y-164 = 0. What is the number of chords with integer chord length?
The diameter of the circle is 2 × 13 = 26
The distance from point a (11,2) to the center of circle (1.2) is 10, so the minimum chord length is 2 √ (13 ^ - 10?) = 2 √ 69 ≈ 16.61
The string is the longest when it passes through the center of the circle, so there are 19 strings with integer length
Given that the diameter of ⊙ o is ab = 10 cm, the chord CD ⊥ AB is at point M. if om: OA = 3:5, what is the length of the string AC?
∵ the diameter of O is ab = 10 cm
OA, OC are the radius of ⊙ o
∴OA=1/2AB=1/2*10=5(cm)
Om: OA = 3:5
So om = 3 / 5 * OA = 3 / 5 * 5 = 3 (CM)
AM=OA-OM=5-3=2(cm)
In the right triangle com, from the Pythagorean theorem, the
CM^2=OC^2-OM^2=5^2-3^2=16
∴CM=4(cm)
In ACM of right triangle, by Pythagorean theorem, we get
AC^2=AM^2+CM^2=2^2+4^2=20
∴AC=2√5(cm)
The length of the string AC is 2 √ 5cm
There is a point P (- 1,2) in the circle (x + 1) ^ 2 + y ^ 2 = 8. AB passes through the point P. if the chord length | ab | = 2 and the root sign 7, find the inclination angle of ab
Passing through the center o as of vertical ab
Then AF = AB / 2 = √ 7, OA = 2 √ 2
So by the Pythagorean theorem of = 1
Is ab, slope = K
y-2=k(x+1)
kx-y+k+2=0
Center (- 1,0)
So of = | - k-0 + K + 2 | / √ (k ^ 2 + 1) = 1
√(k^2+1)=2
k^2=3
So Tana = k = ±√ 3
So the tilt angle is π / 3 or 2 π / 3
As shown in the figure, PA and Pb are tangent to ⊙ o at two points a and B respectively, making diameter AC, and extending the intersection of Pb to point D, connecting OP and CB (1) Confirmation: OP ∥ CB; (2) If PA = 12, DB: DC = 2:1, find the radius of ⊙ o
(1) It is proved that: connect AB, ∵ PA, Pb are tangent to ⊙ o at two points a and B respectively, ᙽ PA = Pb and ᙽ apo = ∵ BPO. ᙽ op ⊥ ab
It is known that ad is the center line of the triangle ABC, e and F are points on AB and AC respectively, and AE = AF, EF intersects ad at point m. It is proved that EM: MF = AC: ab
Make parallel lines of EF from B and C, and intersect with G, h respectively with AD (or extension line)
Because BD = DC, ∠ BDG = ∠ CDH (antiparietal angle) ∠ CHD = ∠ bgd (internal staggered angle of parallel line)
So triangle CDH and triangle BDG are congruent
So BG = ch
EM:BG=AE:AB
MF:CH=AF:AC
Division of two forms
Em: MF = AC: ab can be obtained
As shown in the figure AB is the diameter of circle O, M is the midpoint of inferior arc AC, the chord AC and BM intersect with D, ∠ ABC = 2 ∠ a, it is proved that ad = 2CD emergency
prove:
∵ AB is the diameter of ⊙ o
∴∠C=90°
∵∠ABC=2∠A
∴∠ABC=60°,∠A=30°
∵ m is the midpoint of inferior arc AC
∴∠ABD=∠CBD=30°
∴BD=2CD,∠A=∠ABD
∴AB=BD
∴AD=2CD
In the triangle ABC, ab = 12, points E and D are on AC and ab respectively, AE = 6, EC = 4, AD / DB = AE / EC (1) Find the length of AD; (2) Can dB / AB = EC / AC be established? Please explain the reasons
(1) Because AD / DB = AE / EC = 6 / 4
Ad / DB = 3 / 2,
That is 3dB = 2ad. (1)
Because AB = 12, D is on ab
So 3dB + 2Ab = 12. (2)
Simultaneous equations (1) (2)
The results show that AD=3, DB=9
(2)DB/AB=9/12=3/4
EC/AC=4/10=2/5
So AC / DB is not equal to
So dB / AB = EC / AC does not hold
In the triangle ABC, AB: DB = AC: EC, it is proved that AD: ab = AE: ab ad: DB = AE: EC
Do you want to ask, "De is in triangle ABC respectively AB.AC It is proved that: ad ratio DB = AE to EC, DB = AE to EC, DB to EC, ab = EC to AC
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- 1. It is known that in the triangle ABC, D and E are on AB and AC respectively. If AD / DB = AE / EC, then the following formula holds true 1、AD/EC=DB/AE 2、AB/DB=EC/AE 3、AD/AB=AE/AC 4、AD/DB=BC/EC reason
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