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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
X^2-42X+432=0
（x-18）（x-24）=0
X = 18 or x = 24
If the radius is R and the distance from the center of the circle to the nearest chord is D, then there is
R^2=d^2+12^2
R^2=(d+3)^2+9^2
d=9,R=15
It is known that the radius of center O is 5, AB, CD are two chords of center O, and ab is parallel to CD, ab = 6, CD = 8. Find the distance between AB and CD
It is discussed in two cases
(1) when two parallel lines AB and CD are on the same side of o point of the center of the circle:
The vertical lines of CD and ab are drawn through point O, and the vertical feet are points E and f respectively,
Then EC = ed = 4, FA = FB = 3,
When OA and OC are connected, OA = OC = 5,
According to Pythagorean theorem, OE = 3, of = 4,
∴EF=4－3=1,
The distance between AB and CD is 1 cm;
(2) when the two parallel lines AB and CD are on the opposite side of o point of the circle center:
Similarly, EF = 4 + 3 = 7cm
As shown in the figure, AB is the diameter of
∵AB⊥CD
ν CE = de = 8 (vertical diameter theorem)
And ∵ CE * de = AE * be (intersecting chord theorem)
∴BE=16
∴AB=16+4=20
∴OB=10
∴OE=BE-OB=16-10=6
AB is the circle O diameter CD is the chord AE is perpendicular to e BF is perpendicular to f to prove CE = DF OE = of
prove:
Make om ⊥ CD at point M
∵AE⊥CD,BF⊥CD,OA=OB
∴CM=DM,EM=FM
∴EM-CM=FM-DM
DF = CE
∵ME=MF
/ / OM is the vertical bisector of EF
∴OE=OF
It is known that, as shown in the figure, AB and CD are two chords of circle O, intersecting at point E, AB=CD. verification: 1.OE bisecting [BEC]; 2.AE=DE
(1) Make of ⊥ AB in F, og ⊥ CD in G, ∵ AB = CD, ᙽ of = og (in the same circle, the chord center distance of equal chords is equal), ᚉ EO bisection ᚉ bec (points with equal distance from both sides of an angle are on the bisector of the angle) (2) connect ad, ∵ AB = CD, arc AB = arc CD, ? arc BD = arc AC, ? a = D, AE = De
If the inner point m (1, root 2) is the two mutually perpendicular chords AB and CD of a circle, then the maximum value of AB + CD?
To simplify the calculation, m (1, √ 2) is rotated around o to n (0, √ 3),
If AC: kx-y + √ 3 = 0, then
BD:x+k(y-√3）=0,
The distance from O to AC D1 = (√ 3) / √ (k ^ 2 + 1),
The distance from O to BD is D2 = | K √ 3 | / √ (k ^ 2 + 1),
d1^2+d2^2=3,
(d1d2)^2
A, B, C and D are the four points on circle O. the diameter of circle O is ab = 10, and the chord CD = 8. Then the vertical lines of straight line CD are made through a and B respectively. If the perpendicular feet are m and N, then the quantitative relationship between AM and BM is
Make op ⊥ CD to P through the center O, and connect OC
∵OP⊥CD
∴CP＝CD/2＝8/2＝4
∴OC＝AB/2＝5
∴OP＝√（OC²-CP²）＝√（25-16）＝3
∵AM⊥CD、BN⊥CD
∴AM∥OP∥BN
∵OA＝OB
The OP is the median line of trapezoidal abnm
∴AM+BN＝2OP＝6
The math group answered your question,
It is proved that AB is the chord of circle O and CD is the tangent line passing through a point m on circle O. when AB / / CD, am = BM (2) a AB is the chord of circle O and CD is the tangent passing through a point m on circle o The results showed that: (1) when AB / / CD, am = BM (2) When am = BM, AB / / CD
The first one uses the vertical diameter theorem
The second is also the vertical diameter theorem
It is known that AB is the diameter of ⊙ o, and the chord CD with length of 12cm is vertical AB and vertical foot is m. if om: OA = 3:5, what is the diameter of ⊙ o (why)
If OC is connected, OA = OC (radius of circle). According to Pythagorean theorem (3 strands and 4 strings 5), OM: MC: OC = 3:4:5, so the radius OC = 5 / 4mc = 5 / 4 × 1 / 2 × 12 = 7.5 (CM) 2oC = 15cm, so the diameter is 15cm
We know that the diameter of ⊙ o is ab = 20cm, the chord CD ⊥ AB, and the perpendicular foot is m. if om: OA = 3:5, then the length of string CD is () A.10cm B.12cm C.16cm D.18cm
C
Vertical diameter theorem
AB=20
Radius r = 10 = OA
OM=3/5*10=6
Even OC
Then OC = r = 10
MC = 8.68 8 10 Pythagorean number
CD =2MC =16
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