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Sub 505 (Easy)|   Geometry|                        
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dave13
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The circle with center C shown above is tangent to both axes 02 Mar 2018, 11:18
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Bunuel

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.
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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:

\((x−a)^2+(y−b)^2=r\)
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The circle with center C shown above is tangent to both axes 02 Mar 2018, 12:13
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EMPOWERgmatRichC
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
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Hello EMPOWERgmatRichC, :)

what do you mean by combining 'like' terms before you take the square-root of both sides :? dont understand ... :-)
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dave13
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The circle with center C shown above is tangent to both axes 03 Mar 2018, 13:28
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Bunuel

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.
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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:)
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The circle with center C shown above is tangent to both axes 20 Nov 2022, 12:47
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Walkabout
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) \(\frac{k}{3}\)
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Hi Brent BrentGMATPrepNow, somehow I ended up connecting line from C to X. Would it still works this way? Thanks Brent
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BrentGMATPrepNow
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The circle with center C shown above is tangent to both axes 20 Nov 2022, 13:32
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Kimberly77


Hi Brent BrentGMATPrepNow, somehow I ended up connecting line from C to X. Would it still works this way? Thanks Brent
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I'm not sure what you mean.
"x" isn't a point on the x-axis; the "x" just lets us know that the horizontal line is the x axis.

If you show me your calculations, I'd be happy to take a look.
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The circle with center C shown above is tangent to both axes 21 Nov 2022, 11:36
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BrentGMATPrepNow
Kimberly77


Hi Brent BrentGMATPrepNow, somehow I ended up connecting line from C to X. Would it still works this way? Thanks Brent

I'm not sure what you mean.
"x" isn't a point on the x-axis; the "x" just lets us know that the horizontal line is the x axis.

If you show me your calculations, I'd be happy to take a look.
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Thanks BrentGMATPrepNow. I see what you mean now that I've mistaken X as a point to draw the line to.
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The circle with center C shown above is tangent to both axes 03 Jun 2023, 05:10
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Given that the distance from the origin O to center C is equal to k, we need to find the radius of the circle in terms of k.

Since the circle is tangent to both axes, the radius will be the length of the line segment from the center C to either of the points of tangency on the axes.

Let's consider the point of tangency on the x-axis as A. Since the circle is tangent to the x-axis, the radius and the y-coordinate of point C will be equal.

Now, we have a right triangle OCA, with OC as the hypotenuse (length k) and AC as one of the legs (radius of the circle).

Applying the Pythagorean theorem, we have:

k^2 = AC^2 + OA^2

Since AC is the radius of the circle and OA is the distance from the origin to the point of tangency on the x-axis, which is equal to the radius of the circle as well, we can rewrite the equation as:

k^2 = AC^2 + AC^2
2AC^2 = k^2
AC^2 = k^2 / 2

Taking the square root of both sides, we have:

AC = sqrt(k^2 / 2) = k / sqrt(2)

Therefore, the radius of the circle is equal to AC, which is k divided by the square root of 2.

Hence, the correct answer is (B) k/√2.
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The circle with center C shown above is tangent to both axes 12 Jun 2025, 00:57
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