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

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The circle with center C shown above is tangent to both axes [#permalink] New post 29 Dec 2012, 04:28
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Attachment:
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The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?

(A) k
(B) \frac{k}{\sqrt{2}}
(C) \frac{k}{\sqrt{3}}
(D) \frac{k}{2}
(E) \frac{k}{3}
[Reveal] Spoiler: OA
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Re: The circle with center C shown above is tangent to both axes [#permalink] New post 29 Dec 2012, 04:32
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The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?

(A) k
(B) \frac{k}{\sqrt{2}}
(C) \frac{k}{\sqrt{3}}
(D) \frac{k}{2}
(E) \frac{k}{3}

Look at the diagram below:
Attachment:
Circle2.png
Circle2.png [ 5.4 KiB | Viewed 3090 times ]
Since OC=k, then r^2+r^2=k^2 --> r=\frac{k}{\sqrt{2}}.

Answer: B.
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Re: The circle with center C shown above is tangent to both axes [#permalink] New post 20 Jun 2013, 17:04
How did we conclude that the base of the triangle is also r ?
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Re: The circle with center C shown above is tangent to both axes [#permalink] New post 20 Jun 2013, 17:26
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kkbimalnair wrote:
How did we conclude that the base of the triangle is also r ?



Attachment:
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AC is the radius of the circle and Y-axis is tangent to the circle.Which implies that AC is perpendicular to the Y-axis.

Similarly BC is perpendicular to the X-axis.

Angle AOB is right angle. Which implies angle ACB is also right angled.

Now the quadrilateral OACB must be either a rectangle or a Square.

In either case OB must be equal to AC(as both in a rectangle and in a square opposite sides are equal) . So OB = R
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Re: The circle with center C shown above is tangent to both axes [#permalink] New post 04 Nov 2013, 16:30
I actually solved this one with a bit of logic, or at least eliminated several answer choices.

A) is obviously out since k includes that area outside of the circle by the origin, so the radius must be shorter. Dividing it by 2 or 3 seemed overkill. dividing k by the sqrt of 3 implied some sort of 30/60/90 triangle which didn't make much sense. so I figured it was B, then made the triangle real quick, saw it was a right triangle and as soon as I wrote out the pythag equation I saw that B was definitely correct.
Re: The circle with center C shown above is tangent to both axes   [#permalink] 04 Nov 2013, 16:30
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