As shown in the figure, the radii of circles O1 and O2 are 5cm and 4cm respectively. The two circles intersect points a and B, and ab = 6cm. Find the length of the distance between the center of the circle and O1O2

As shown in the figure, the radii of circles O1 and O2 are 5cm and 4cm respectively. The two circles intersect points a and B, and ab = 6cm. Find the length of the distance between the center of the circle and O1O2

Let the intersection of AB and O! O2 be c
Because AB is perpendicular to O1O2
Easy to get according to Pythagorean theorem
O1C=4
O2C = radical 7
Therefore, the center distance is 4 + radical 7

As shown in the figure, in the equilateral △ ABC with side length L, the circle O1 is the inscribed circle of △ ABC, the circle O2 is circumscribed with circle O1, and tangent to AB, BC The circle on + 1 is circumscribed to the circle on and tangent to AB, BC, and so on. Let's say that the area of the circle on is an (n ∈ n) (I) it is proved that {an} is an equal ratio sequence; (II) find Lim n→∞（a1+a2+… +Value an)

(I) prove that RN is the radius of the circle on,
Then R1 = L
2tan30°＝
Three
6l，
rn−1−rn
rn−1+rn＝sin30°＝1
2．
So RN = 1
3rn−1(n≥2)，
So A1 = π R
One
Two
＝πl2
12，
An
an−1＝(rn
rn−1)2＝1
Nine
Therefore {an} is a sequence of proportional numbers
(II) because an = (1)
9)n−1a1(n∈N)，
So Lim
n→∞(a1+a2+… +an)＝a1
1−1
9＝3πl2
32．

In the equilateral triangle ABC of length L, the circle O1 is the inscribed circle of △ ABC, and the circle O2 is circumscribed with O1 and tangent to AB and BC The circle on + 1 is tangent to on, and tangent to AB and BC. If this continues indefinitely, let's say that the area of circle on is an (n ∈ n *) General term formula of sequence {an}

The tangent line of circle O2 and O1 cuts out a small equilateral triangle EBF whose side length is L / 3
The radius of O1 = R1 = (√ 3 / 6) l [please also prove by the owner]. [radius of O2 = (√ 3 / 6) (L / 3)
Radius of on = RN = (√ 3 / 6) (L / 3 ^ (n-1))
The area of on is an = π (RN) 2 = π L 2 / [4x3 ^ (2n-1)]

It is known that: as shown in the figure, circle O1 and circle O2 intersect at two points of AB, and the center of circle O1 is on circle O2, the diameter of circle O2 AC intersects circle O1 and point D, and the extension line of CB intersects circle O1 at e, indicating that ad = be

Connect AB in ⊙ O2,
∵ AC is the diameter
∴∠ABC=90°,∠ABE=90°
In ⊙ O1, connect AE and ed
∵∠ABE=90°
ν AE is the diameter, O1 point is on AE, ∠ EDA = 90 °
Connect CO1,
∵ O1 point on ⊙ O2
∴∠CO1A=90°,
And ∵ AO1 = o1e
/ / CO1 is the vertical bisector of AE,
Then CE = Ca, ∠ CEA = ∠ CAE;
In RT △ EDA and RT △ Abe,
AE=AE
∠BEA=∠DAE,
∴Rt△EDA≌Rt△ABE,
∴AD=BE.

As shown in the figure, circle O2 and semicircle O1 are inscribed at C, and tangent to the diameter ab of radius and point D. if AB = 6, the radius of circle O2 is 1. Find ∠ ABC

As long as we can find out the degree of the angle co1b, we can use the circumference angle of the same chord circle as half of the center angle to solve the problem
Firstly, because the radius of O2 is 1 and the radius of O1 is 3, we can know that o2d = 1 o2o1 = co1-co2 = 2
O1b is tangent to O2, so o2d is perpendicular to o1b, so we can know that the degree of angle co1b is 30 ° then the degree of angle cab is 15 ° and the degree of angle CBA is 90 ° - 15 ° = 75 °

It is known that the diameter of circle O is ab = 8, the radius OC is perpendicular to AB, and OC is the diameter of circle O1. Circle O2 is inscribed with circle O, circumscribed with circle O1 and tangent to ab respectively. It is proved that ⊙ O1 is tangent to ab and⊙ o respectively

Well, I'd like to ask the next question

Known: as shown in the figure, ⊙ O1 and ⊙ O2 intersect at a and B, point O2 is on ⊙ O1, ad is the diameter of ⊙ O2, connect dB and extend ⊙ O1 to point C Verification: 1 2AD＝ CD2−CO22．

Proof: connect ab,
In △ bad and △ co2d
∵∠BAD=∠C，∠D=∠D，
∴∠ABD=∠CO2D，
∵ ad is ⊙ O2 diameter,
∴∠ABD=90°=∠CO2D，
In RT △ co2d, o2d=
CD2−CO
Two
Two
,
And ∵ o2d = 1
2AD，
∴1
2AD=
CD2−CO
Two
Two
.

⊙ O1 and ⊙ O2 are circumscribed at point P, the straight line AB passing through point P intersects with ⊙ O1 and ⊙ O2 at a and B, and the tangent ad of ⊙ O1 intersects with ⊙ O2 at points c and D BC= BD．

It is proved that, as shown in the figure, the common tangent Mn of two circles is made through point P. BD, PD and CB are connected
∵ ad, ⊙ O1 tangent, Mn is the common tangent,
∴∠1=∠2=∠3=∠6．
∵∠4=∠5，∠BDC=∠5+∠6，∠BCD=∠1+∠4，
∴∠BDC=∠BCD，
Qi
BC=
BD．

As shown in the figure, circle O1 intersects circle O2 at two points ab. circle O1 is on circle O2, chord BC of circle O2 cuts circle O1 at point B, extends Bo1, CA is called intersection with P, Pb and circle O1 and D Find that AC is the tangent of circle O1

Link AO1
∵ BC cuts ⊙ O1 to B,  cbo1 = 90 °
∵ ao1bc is the inscribed quadrilateral of a circle, ᙽ pao1 ᙽ cbo1 = 90 °, and

As shown in the figure, ⊙ O1 and ⊙ O2 intersect at a and B, O1 is on ⊙ O2, chord BC of ⊙ O2 cuts ⊙ O1 at B, extends Bo1 and Ca, intersects point P, Pb and ⊙ O1 at point D (1) O1 is proved to be tangent; (2) Connect AD and o1c, confirm that ad ∥ o1c; (3) If PD = 1, the radius of ⊙ O1 is 2, find the length of BC

(1) It is proved that: connect O1A; ∵ BC is tangent of ⊙ O1,  o1bc = 90 °. ∵ o1ap is the outer angle of inscribed quadrilateral of circle O2,  pao1 = ∠ o1bc = 90 ° and  Q1A ⊥ AC, then AC is tangent of ⊙ O1. (2) it is proved that: connect AB, ∵ PC cut ⊙ O1 at point a, ∵ pad = ∵ abd