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

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

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New post 02 Mar 2018, 11:18
Bunuel wrote:
Image
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
Since OC=k, then \(r^2+r^2=k^2\) --> \(r=\frac{k}{\sqrt{2}}\).

Answer: B.



Dear Bunuel, hello ! :-)

i know you are tired of my questions :) but i am sure you still can understand me :)

can you please explan why do you use this formula? \(r^2+r^2=k^2\)

from this post https://gmatclub.com/forum/math-coordin ... 87652.html

i know that the if the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2 = r^2\) and in other cases

we use this formula In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:

\((x−a)^2+(y−b)^2=r\)
Director
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 02 Mar 2018, 12:13
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


Hello EMPOWERgmatRichC, :)

what do you mean by combining 'like' terms before you take the square-root of both sides :? dont understand ... :-)
Director
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The circle with center C shown above is tangent to both axes  [#permalink]

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New post 03 Mar 2018, 13:28
Bunuel wrote:
Image
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
Since OC=k, then \(r^2+r^2=k^2\) --> \(r=\frac{k}{\sqrt{2}}\).

Answer: B.


Bunuel shouldnt we rationalise denominator :) \(r=\frac{k}{\sqrt{2}}\).

for instance if we have \(\frac{\sqrt{6}}{\sqrt{2}}\) = \(\frac{\sqrt{6}}{\sqrt{2}}\) *\(\frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{\sqrt{12}}{\sqrt{4}}\)
so we get \(\frac{\sqrt{12}}2\)

As you see in the denominator 2 is without radical sign, so why you didn't write it so ?

Another question sqrt of 2 is 1.4 you could write it so in the denominator too ? in which case you could write 1.4 ? can you please explain ? :) pleaseee:)
GMAT Club Bot
The circle with center C shown above is tangent to both axes &nbs [#permalink] 03 Mar 2018, 13:28

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