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VP  D
Joined: 09 Mar 2016
Posts: 1230
The circle with center C shown above is tangent to both axes  [#permalink]

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Bunuel wrote: 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 $$r^2+r^2=k^2$$ --> $$r=\frac{k}{\sqrt{2}}$$.

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:

$$(x−a)^2+(y−b)^2=r$$
VP  D
Joined: 09 Mar 2016
Posts: 1230
Re: The circle with center C shown above is tangent to both axes  [#permalink]

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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

Hello EMPOWERgmatRichC, what do you mean by combining 'like' terms before you take the square-root of both sides dont understand ... VP  D
Joined: 09 Mar 2016
Posts: 1230
The circle with center C shown above is tangent to both axes  [#permalink]

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Bunuel wrote: 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 $$r^2+r^2=k^2$$ --> $$r=\frac{k}{\sqrt{2}}$$.

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:)
Non-Human User Joined: 09 Sep 2013
Posts: 13157
Re: The circle with center C shown above is tangent to both axes  [#permalink]

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_________________ Re: The circle with center C shown above is tangent to both axes   [#permalink] 16 Mar 2019, 15:29

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