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Re: In the figure, P and Q are centers of the two circles of rad [#permalink]
Carcass wrote:
Attachment:
#GREpracticequestion In the figure, P and Q are centers .jpg


In the figure, P and Q are centers of the two circles of radii 3 and 4, respectively. A and B are the points at which a common tangent touches each circle.

Quantity A
Quantity B
\(AB\)
\(PQ\)



Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.


I don't understand. A well-detailed explanation would be appreciated.
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Re: In the figure, P and Q are centers of the two circles of rad [#permalink]
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Then ABPD must be a rectangle. Hence, AB = DP. Also, ∠PDQ equals the corresponding angle, ∠BAD.So, both equal 90°. Since a right angle is the greatest angle in a triangle, the side opposite the angle is the longest. PQ is the side opposite the right angle ∠PDQ in ∆PQD. Hence, PQ is greater than the other side DP. Hence, PQ is also greater than AB, which equals DP (we know). Hence, Column B is greater than Column A, and the answer is (B).
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Re: In the figure, P and Q are centers of the two circles of rad [#permalink]
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Another way to solve this is

PQ is distance between two centres and since they are touching the distance will be sum of radii which is 7
AB is a direct common tangent from same external point to both the circles

so using direct common tangent formulae which is ((distance between centre)^2-(radii 1-radii2)^2)^1/2.
So AB=((49-(1))^1/2=48^1/2=6.xx
So PQ>AB
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Re: In the figure, P and Q are centers of the two circles of rad [#permalink]
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