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## geometry circle

This topic has 1 expert reply and 2 member replies
chamisool Junior | Next Rank: 30 Posts
Joined
09 Sep 2011
Posted:
11 messages
1

#### geometry circle

Thu Oct 27, 2011 8:19 am
ttp://postimage.org/image/81u46oy13/" target="_blank">

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) k/(sqrt 2)
(C) k/(sqrt 3)
(D) k/2
(E) k/3

Can anyone help me with this question? I understand the answer but the answer I originally got was not in the answer choices.

What I did was I extended k past the center of the circle, so I form the circle inscribed in a square.

As k is double the length, I made it so its 2k as the hypotenuse. The ratio formula for an equilateral right triangle is x : x : x*sqrt 2
So I did x*sqrt 2 = 2k
Isolate x and you get x = 2k/(sqrt 2)

So the legs of the triangle is 2k/(sqrt 2). I simplified it by multiplying both numerator and denominator by sqrt 2, so I got
X = (2k*sqrt 2)/2 which becomes k*sqrt 2. Since the leg is the diameter, I divided it by 2 and my final answer was (k*sqrt 2)/2 but the answer choice isnâ€™t there.

Is it wrong to simplify 2k/(sqrt 2)? Because if I get 2k/sqrt 2 divided by 2, then I get the correct answer. I guess my question is when do you leave the sqrt denominator alone and when do you multiply both numerator and denominator by the sqrt?

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chamisool Junior | Next Rank: 30 Posts
Joined
09 Sep 2011
Posted:
11 messages
1
Thu Oct 27, 2011 8:55 am
Amiable Scholar wrote:
the side of your square is not the radius but equal to diameter of the circle.
so you need to divide it by 2
(2k/sqrt(2))/2
=k/sqrt(2)
Yeah, that is what I did. But before I divided the diameter by 2 I multiplied the denominator and numerator by sqrt 2 to simply my answer.

So I got 2k/(sqrt 2) *(sqrt 2/sqrt 2)

So I got (2k*sqrt 2)/2 = k*sqrt 2

So I got diameter of k*sqrt 2 and then divided that by 2 to find the radius. (k*sqrt 2)/2

I guess I'm confusing myself cause I noticed based Anurag's response, if I multiply the numerator and denominator again, the answer becomes k/sqrt 2.

### GMAT/MBA Expert

Anurag@Gurome GMAT Instructor
Joined
02 Apr 2010
Posted:
3835 messages
Followed by:
519 members
1854
GMAT Score:
770
Thu Oct 27, 2011 8:28 am
chamisool wrote:
... my final answer was (k*sqrt 2)/2 but the answer choice isnâ€™t there.
2 = (âˆš2)*(âˆš2)

Hence, âˆš2/2 = 1/âˆš2
Hence, kâˆš2/2 = k/âˆš2 --> Option B

As far as solving the problem is concerned, there is no need to extend k and bring complications. Just draw a perpendicular from C on the x-axis. Then you will get an isosceles right angled triangle with hypotenuse of length k.

Hence, length of the other sides = length of radius = k/âˆš2

_________________
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

GMAT with Gurome

Thanked by: chamisool
chamisool Junior | Next Rank: 30 Posts
Joined
09 Sep 2011
Posted:
11 messages
1
Thu Oct 27, 2011 8:55 am
Amiable Scholar wrote:
the side of your square is not the radius but equal to diameter of the circle.
so you need to divide it by 2
(2k/sqrt(2))/2
=k/sqrt(2)
Yeah, that is what I did. But before I divided the diameter by 2 I multiplied the denominator and numerator by sqrt 2 to simply my answer.

So I got 2k/(sqrt 2) *(sqrt 2/sqrt 2)

So I got (2k*sqrt 2)/2 = k*sqrt 2

So I got diameter of k*sqrt 2 and then divided that by 2 to find the radius. (k*sqrt 2)/2

I guess I'm confusing myself cause I noticed based Anurag's response, if I multiply the numerator and denominator again, the answer becomes k/sqrt 2.

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