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The circle with center C shown above is tangent to both axes and has

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The circle with center C shown above is tangent to both axes and has  [#permalink]

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New post 22 Aug 2019, 01:19
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The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?

A. A

B. \(\sqrt{\frac{2A}{\pi}}\)

C. \(\sqrt{\frac{3A}{\pi}}\)

D. \(\frac{2A}{\pi}\)

E. \(\frac{3A}{\pi}\)


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Re: The circle with center C shown above is tangent to both axes and has  [#permalink]

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New post 22 Aug 2019, 01:19
Bunuel wrote:
The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?

A. A

B. \(\sqrt{\frac{2A}{\pi}}\)

C. \(\sqrt{\frac{3A}{\pi}}\)

D. \(\frac{2A}{\pi}\)

E. \(\frac{3A}{\pi}\)


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Re: The circle with center C shown above is tangent to both axes and has  [#permalink]

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New post 22 Aug 2019, 01:36
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Hi Bunuel,

Should there be an attachment?
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Re: The circle with center C shown above is tangent to both axes and has  [#permalink]

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Re: The circle with center C shown above is tangent to both axes and has  [#permalink]

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New post 22 Aug 2019, 01:55
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Draw perpendiculars from C to the x and y axes. We now have a square with side \(r\)(radius of circle) and diagonal \(OC=\sqrt{2}r\)

We know that Area, \(A=\pi r^2\) and so \(r=\sqrt{\frac{A}{\pi}}\)

So \(OC=\sqrt{2}*\sqrt{\frac{A}{\pi}}=\sqrt{\frac{2A}{\pi}}\)

Answer is (B)

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Re: The circle with center C shown above is tangent to both axes and has  [#permalink]

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New post 22 Aug 2019, 04:42
let center from x axis be (3,3) so distance o to c ; 3√2
plug in a = 9pi
we get B is correct
IMO B

Bunuel wrote:
Image
The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?

A. A

B. \(\sqrt{\frac{2A}{\pi}}\)

C. \(\sqrt{\frac{3A}{\pi}}\)

D. \(\frac{2A}{\pi}\)

E. \(\frac{3A}{\pi}\)


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Attachment:
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Re: The circle with center C shown above is tangent to both axes and has  [#permalink]

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New post 24 Aug 2019, 05:28
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Bunuel wrote:
Image
The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?

A. A

B. \(\sqrt{\frac{2A}{\pi}}\)

C. \(\sqrt{\frac{3A}{\pi}}\)

D. \(\frac{2A}{\pi}\)

E. \(\frac{3A}{\pi}\)





Attachment:
GFPS2110.jpg


Since the area is A, we have:

A = πr^2

A/π = r^2

√(A/π) = r

Note that if we drop a perpendicular from C to the x axis (call it B), then we have created triangle OCB that is a 45-45-90 triangle with legs r and hypotenuse r√2 (which is also the distance from O to C). Recall that the side ratio for a 45-45-90 triangle is x : x : x√2 .

Since the radius is √(A/π), the distance from C to O is √(A/π) * √2 = √(2A/π).

Answer: B
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Re: The circle with center C shown above is tangent to both axes and has   [#permalink] 24 Aug 2019, 05:28
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