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The circle at left above is tangent to the x [#permalink]
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\(\frac{1}{d-c}>\frac{1}{7}\)


\(\frac{1}{d-c}=\frac{1}{4}\)
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Re: The circle at left above is tangent to the x [#permalink]
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Carcass wrote:
Attachment:
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\(\frac{1}{d-c}>\frac{1}{7}\)


\(\frac{1}{d-c}=\frac{1}{4}\)


Ah, I see my mistake now. Thank you!
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The circle at left above is tangent to the x [#permalink]
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Carcass Ks1859

Guys this one is bit confusing!
The centre of the bigger circle will not be at (10, 4)

Why?
Because, the distance between the 2 centres must be 7 right?! (3 + 4)

But, if you apply Distance formula between (3, 3) and (10, 4)
Distance = \(\sqrt{(4-3)^2 + (10-3)^2} = \sqrt{50} = 7.07\)

Which is not possible!

I think the figure provided by Carcass is more than sufficient to answer the question. Even though my answer was correct but my explanation was still at fault!
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Re: The circle at left above is tangent to the x [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep 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: The circle at left above is tangent to the x [#permalink]
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