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# If 2 equal circles of radius 5cm have 2 common tangents AB and CD whic

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Joined: 19 Aug 2009
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If 2 equal circles of radius 5cm have 2 common tangents AB and CD whic  [#permalink]

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02 Oct 2009, 10:16
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Difficulty:

55% (hard)

Question Stats:

57% (02:31) correct 43% (02:00) wrong based on 22 sessions

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If 2 equal circles of radius 5cm have 2 common tangents AB and CD which touch the circle on A,C and B,D respectively and if CD= 24 cm, find the length of AB.

A. 27cm
B. 25cm
C. 26cm
D. 30cm

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Joined: 02 Sep 2009
Posts: 52288
If 2 equal circles of radius 5cm have 2 common tangents AB and CD whic  [#permalink]

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04 Oct 2009, 13:13

If 2 equal circles of radius 5cm have 2 common tangents AB and CD which touch the circle on A,C and B,D respectively and if CD= 24 cm, find the length of AB.

A. 27cm
B. 25cm
C. 26cm
D. 30cm

Two line interception - K. Center of first circle O, of second P.

OC/KC = (KC + CD)/PD --> OC = PD = R = 5 --> 5/KC = (24 + KC)/5 --> Solving for KC --> KC = 1.

KC = AK = 1, KD = KB = KC + CD = 1 + 24 = 25 --> AB = AK + KB = 1 + 25 = 26.

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Re: If 2 equal circles of radius 5cm have 2 common tangents AB and CD whic  [#permalink]

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16 Sep 2010, 21:18
Bunuel wrote:
Two line interception - K. Center of first circle O, of second P.

OC/KC=(KC+CD)/PD --> OC=PD=R=5 --> 5/KC=(24+KC)/5 --> Solving for KC --> KC=1.

KC=AK=1, KD=KB=KC+CD=1+24=25 --> AB=AK+KB=1+25=26

Bunuel, I did not understand your explanation. Can you please be a bit elaborate?

Specifically, not sure of how you arrived at -- OC/KC = (KC+CD)/PD and KC = AK = 1.
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Re: If 2 equal circles of radius 5cm have 2 common tangents AB and CD whic  [#permalink]

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16 Sep 2010, 22:15
Take the intersection of line AB and CD as point E.

EB = ED as tangents from a single point to a circle are equal

EB = EC + CD take EC = x
=> EB = x+ 24

Also AE=EC = x using same explanation above.

AB = AE+EB = x+x+24 = 2x+24 = even number- if you are short of time you can easily guess.

Now triangle ECO is similar to triangle triangle EDO' where O and O' are center of 1st and 2nd respectively.

=>$$\frac{EC}{OD} = \frac{O'D}{ED}$$

=> $$\frac{x}{5}= \frac{5}{(x+24)}$$

=> $$x^2 +24x -25 = 0$$
=> $$(x +25)*(x-1)= 0$$ => $$x =1$$

=> AB$$= 2x*+24 = 2+24 = 26$$
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Re: If 2 equal circles of radius 5cm have 2 common tangents AB and CD whic  [#permalink]

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21 Sep 2018, 11:14
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If 2 equal circles of radius 5cm have 2 common tangents AB and CD whic &nbs [#permalink] 21 Sep 2018, 11:14
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