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

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29 Dec 2012, 04:28

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A

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C

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E

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87% (02:15) correct
13% (01:15) wrong based on 960 sessions

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Attachment:

Circle.png [ 3.73 KiB | Viewed 16752 times ]

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) \(\frac{k}{3}\)

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 14769 times ]

Since OC=k, then \(r^2+r^2=k^2\) --> \(r=\frac{k}{\sqrt{2}}\).

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

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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]

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31 Dec 2014, 06:14

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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...

Radius = 2

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.

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

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09 Feb 2016, 12:47

Could someone explain me why I cannot take the square root of this entire expression: \(\sqrt{k^2 = r^2 + r^2}\) ---> \(k = r + r.\) Why is this incorrect?

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

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10 Feb 2016, 04:20

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EMPOWERgmatRichC wrote:

Hi saiesta,

You have to combine 'like' terms before you take the square-root of both sides.

Here's a simple example that proves WHY:

\(\sqrt{4}\) = 2

\(\sqrt{(2+2)}\) does NOT = \(\sqrt{2}\) + \(\sqrt{2}\) though

\(\sqrt{2}\) = about 1.4

2 does NOT equal 1.4 + 1.4

Knowing THAT, how would you write your equation now?

GMAT assassins aren't born, they're made, Rich

Thank you, I have to remember myself to combine like terms before I take the square root of both sides. I believe this is particularly the case when adding or subtracting (square) roots. I would write the equation as follows: \(k^2 = r^2 + r^2\) ----> \(k^2 = 2r^2\) ----> \(\frac{k^2}{2} = r^2\) ----> \(\sqrt{\frac{k^2}{2}} = \sqrt{r^2}\) ----> \(\frac{k}{\sqrt{2}} = r\)

Nicely done. As a minor aside, since this is a Geometry question, you don't have to worry about any of the 'measures' ending up negative. However, if the same concepts were in an Algebraic prompt, then you'd have to keep in mind that both K and R could be negative (so square-rooting both sides might end up eliminating possible answers; in a DS question, that could lead to a wrong answer).

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

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30 Apr 2016, 18:58

Walkabout wrote:

Attachment:

Circle.png

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) \(\frac{k}{3}\)

If we draw radius to two point of tangency(we will have two 90 degrees there and other two angles will also be 90 degrees. find out why.) then we will draw a square with 4 sides equal(find out why ). Now we have diagonal of square = k . then side = k/sqrt2.(why?= 45-45-90 right angle triangle.)

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

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02 May 2016, 08:20

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Walkabout wrote:

Attachment:

Circle.png

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) \(\frac{k}{3}\)

We begin by creating a right triangle with the x-axis, the radius of circle C to the x-axis and the line segment from center of circle C to the origin. We see that we have created an isosceles right triangle, also known as a 45-45-90 degree right triangle. We know this because each leg of the right triangle is equal to a radius of the circle. We can label all this in our diagram.

We know the side-hypotenuse ratios in a 45-45-90 degree right triangle are:

x:x:x√2, where x represents the leg of the triangle and x√2 is the hypotenuse.

We can use this to determine the leg of the triangle.

Since x√2 equals the hypotenuse of the triangle we can say:

x√2 = k

x = k/√2

Since x also represents the radius of the circle, k/√2 is equal to the radius.

The answer is B.
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