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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]  29 Dec 2012, 04:28
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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]  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 [ 5.4 KiB | Viewed 8400 times ]
Since OC=k, then $$r^2+r^2=k^2$$ --> $$r=\frac{k}{\sqrt{2}}$$.

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

Untitled.jpg [ 6.15 KiB | Viewed 7767 times ]

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]  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.
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Re: The circle with center C shown above is tangent to both axes [#permalink]  31 Dec 2014, 06:14
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Re: The circle with center C shown above is tangent to both axes [#permalink]  31 Dec 2014, 11:09
Expert's post
Hi All,

This question can be solved by TESTing VALUES.

First though, there's a hidden pattern worth noting: Since the circle is tangent to both the x-axis and the y-axis, we can draw a SQUARE into the picture (using the Origin as one corner and "C" as the opposite corner). The length of a side of this square will EQUAL the RADIUS of the circle. From here, we can make the dimensions of the square anything that we want.

Let's TEST...

The length from the Origin to the center C cuts through the square and forms two 45/45/90 right triangles. Thus...

K = That length = 2\sqrt{2}

Now we just have to plug THAT value into the answers to find the one that equals 2.

The only answer that matches is
[Reveal] Spoiler:
B

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Re: The circle with center C shown above is tangent to both axes   [#permalink] 31 Dec 2014, 11:09
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# The circle with center C shown above is tangent to both axes

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