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Math Expert V
Joined: 02 Sep 2009
Posts: 58117
The circle with center C shown above is tangent to both axes and has  [#permalink]

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Difficulty:   5% (low)

Question Stats: 89% (01:24) correct 11% (02:09) wrong based on 46 sessions

### HideShow timer Statistics 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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Math Expert V
Joined: 02 Sep 2009
Posts: 58117
Re: The circle with center C shown above is tangent to both axes and has  [#permalink]

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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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Hi Bunuel,

Should there be an attachment?
Math Expert V
Joined: 02 Sep 2009
Posts: 58117
Re: The circle with center C shown above is tangent to both axes and has  [#permalink]

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firas92 wrote:
Hi Bunuel,

Should there be an attachment?

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

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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 $$OC=\sqrt{2}*\sqrt{\frac{A}{\pi}}=\sqrt{\frac{2A}{\pi}}$$

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

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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: 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:
GFPS2110.jpg

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

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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. $$\frac{3A}{\pi}$$

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).

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. 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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