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The center distance of the two strings is 3 and 4 respectively
When they are on the same side of the center of the circle, the distance is 4-3 = 1
When they are on the opposite side of the circle center, the distance is 4 + 3 = 7

If the radius of ⊙ o is 5cm, the chord ab ∥ CD, ab = 6cm, CD = 8cm, then the distance between AB and CD is () A. 1 cm B. 7 cm C. 1 cm or 7 cm D. It's impossible to judge

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There are two cases: ① when AB and CD are on the same side of O, as shown in Figure 1,
O as OE ⊥ AB in E, Cd in F, OA, OC,
∵AB∥CD，
∴OF⊥CD，
According to the vertical diameter theorem, AE = 1
2AB=3cm，CF=1
2CD=4cm，
In RT △ OAE, from Pythagorean theorem, OE is obtained=
OA2−AE2=
52−32=4（cm）
Similarly, of = 3cm is obtained,
EF=4cm-3cm=1cm；
II.
When AB and CD are on both sides of O, as shown in Fig. 2, OE = 4cm, of = 3cm can be obtained by the same method,
Then EF = 4cm + 3cm = 7cm;
The distance between AB and CD is 1cm or 7cm,
Therefore, C

In the circle O, the diameter AB = 10cm, the chord AC = 6cm, the bisector of ∠ ACB intersects the circle O at point D, and the lengths of BC, ad, and BD are obtained

Therefore, from the Pythagorean theorem, we know that: the square of BC = the square of ab - the square of AC, then: BC = 8 ﹣ x0d, because the bisector of ﹣ ACB intersects the circle O at point D, ﹤ ACD ﹤ DCB ﹣ x0d, so ad = dB (the same circle circumference angle is the corresponding chord and arc are equal) x0d, so the triangle AB

In the circle O, the diameter AB = 10cm, the chord AC is 8cm, the bisector of ∠ ACB intersects circle O and point D, then BC = how much, AD + BD equals

 diameter ab  ACB = 90,  ADB = 90  BC  BC = √ (AB ∵ AC ? 100-64) = 6 (CM) ? CD bisection  ACB  ACD ? BCD ? ACB / 2 = 45 ∵ abd, ∠ ACD are all inferior arcs ad, and the arcs corresponding to ﹤ bad and ∠ BCD are all inferior arcs BD ﹤ abd ﹤ a

As shown in the figure, the diameter ab of ⊙ o is 10cm, the chord AC is 6cm, the bisector of ⊙ ACB intersects ⊙ o in D, and find the length of BC, ad, BD

 AB is the diameter  ACB = ∠ ADB = 90 ° in RT △ ABC, AB2 = ac2 + BC2, ab = 10cm, AC = 6cm ᙽ BC2 = ab2-ac2 = 102-62 = 64 ᙽ BC = 64 = 8 (CM) and CD bisection

As shown in the figure, ⊙ O's diameter AB bisects chord CD, CD = 10 cm, AP: Pb = 1:5

Connect Co, let the radius of the circle be r, ∵ diameter AB bisecting chord CD, ᙽ AB vertical CD (2 points) ∵ AP: Pb = 1:5, ᙽ if AP = k, Pb = 5K, then AB = AP + Pb = 6K, ᙽ OA = 3k, Po = oa-ap = 3k-k = 2K, ᙽ Po = 23oa = 23R (3) R2 = 52 + (23R) 2, R2 = 45

As shown in the figure, the radius of ⊙ o is 5cm, P is the outer point of ⊙ o, Op = 8cm. Make a circle with P as the center of the circle and circumscribe ⊙ o, and the radius of the circle is 1. As shown in the figure, the radius of ⊙ o is 5cm, P is the outer point of ⊙ o, Op = 8cm. Take P as the center of the circle to make a circle and ⊙ o circumscribed. What is the radius of the circle? 2. Given that ⊙ A is tangent to ⊙ B, ab = 10cm, where the radius of ⊙ A is 4cm, calculate the radius of ⊙ B 5. As shown in the figure, two equal circles ⊙ O and ⊙ o ′ intersect at two points P and Q, TP and NP are tangent lines of ⊙ O and ⊙ o 'passing through point P respectively, and calculate the size of ⊙ TPN

(1) The radius of circumscribed circle is 3cm and that of inscribed circle is 13cm
(2) Compare the radius of B by 6cm or 10cm

It is known that if the secant PAB of ⊙ o intersects ⊙ o at points a, B, PA = 7cm, ab = 5cm, Po = 10cm, then the radius of ⊙ o is () A. 4cm B. 5cm C. 6cm D. 7cm

Extend Po to meet D,
∵PA=7cm，AB=5cm，
∴PB=12cm；
Let the radius of the circle be X,
∵PA•PB=PC•PD，
∴（10-x）（10+x）=84，
∴x=4．
Therefore, a

As shown in the figure, AB is the diameter of ⊙ o, the chord CD intersects AB at P, and PA = 1cm, Pb = 5cm, ∠ DPB = 30 ° and M is the midpoint of CD, and find the length of OM

∵ m is the midpoint of CD and ab is the diameter of ⊙ o
∴OM⊥CD
∵PA＝1cm,PB＝5cm
∴OA＝3cm,∴OP＝2cm
In RT △ POM, ∠ DPB = 30 °
∴OM=1/2OP=1/2×2=1cm