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

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

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New post 29 Dec 2012, 04:28
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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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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 29 Dec 2012, 04:32
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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
Circle2.png [ 5.4 KiB | Viewed 39639 times ]
Since OC=k, then \(r^2+r^2=k^2\) --> \(r=\frac{k}{\sqrt{2}}\).

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

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New post 20 Jun 2013, 17:26
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kkbimalnair wrote:
How did we conclude that the base of the triangle is also r ?



Attachment:
Untitled.jpg
Untitled.jpg [ 6.15 KiB | Viewed 38950 times ]


AC is the radius of the circle and Y-axis is tangent to the circle.Which implies that AC is perpendicular to the Y-axis.

Similarly BC is perpendicular to the X-axis.

Angle AOB is right angle. Which implies angle ACB is also right angled.

Now the quadrilateral OACB must be either a rectangle or a Square.

In either case OB must be equal to AC(as both in a rectangle and in a square opposite sides are equal) . So OB = R
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 20 Jun 2013, 17:04
How did we conclude that the base of the triangle is also r ?
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 04 Nov 2013, 16:30
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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.
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 31 Dec 2014, 11:09
Hi All,

This question can be solved by TESTing VALUES.

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.

The only answer that matches is

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

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New post 04 Dec 2015, 07:10
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alternate method, similar to using the right angle hypotenuese formula:

notice that the figure also forms a cube with r (radius) as each side, and diagonal is k or CO.
Using diagonal of square formula
r√2 = k
r = k/√2

(diagonal of a square is x√2)
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 07 Dec 2015, 05:57
Mantis wrote:
kkbimalnair wrote:
How did we conclude that the base of the triangle is also r ?



Attachment:
Untitled.jpg


AC is the radius of the circle and Y-axis is tangent to the circle.Which implies that AC is perpendicular to the Y-axis.

Similarly BC is perpendicular to the X-axis.

Angle AOB is right angle. Which implies angle ACB is also right angled.

Now the quadrilateral OACB must be either a rectangle or a Square.

In either case OB must be equal to AC(as both in a rectangle and in a square opposite sides are equal) . So OB = R



Hi Can you please let me know how you made AOB as right angle?
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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

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New post 09 Feb 2016, 18:59
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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?

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

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New post 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?

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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: \(k^2 = r^2 + r^2\) ----> \(k^2 = 2r^2\) ----> \(\frac{k^2}{2} = r^2\) ----> \(\sqrt{\frac{k^2}{2}} = \sqrt{r^2}\) ----> \(\frac{k}{\sqrt{2}} = r\)
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 10 Feb 2016, 09:38
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Hi saiesta,

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).

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

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New post 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.)
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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 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.

Image

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

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New post 04 Oct 2016, 23:26
sin45=r/k
1/√2 = r/k
r=k/√2
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 18 May 2017, 08:09
Based on Pythagorean theorem,
r = k/\(\sqrt{2}\)

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

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New post 30 May 2017, 15:26
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Attached is a visual that should help.
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 04 Sep 2017, 10:37
Mantis wrote:
kkbimalnair wrote:
How did we conclude that the base of the triangle is also r ?



Attachment:
Untitled.jpg


AC is the radius of the circle and Y-axis is tangent to the circle.Which implies that AC is perpendicular to the Y-axis.

Similarly BC is perpendicular to the X-axis.

Angle AOB is right angle. Which implies angle ACB is also right angled.

Now the quadrilateral OACB must be either a rectangle or a Square.

In either case OB must be equal to AC(as both in a rectangle and in a square opposite sides are equal) . So OB = R




Bunuel, abhimahna, and experts

Is my understanding of this question correct ?

As two tangents drawn from the same point are equal, then OA=OB.

Radius AC and CB are perpendicular to the two tangents and make 90 degree angle.

Angle AOB is also 90 degrees because y axis is perpendicular to x axis.

So angle ACB is also 90 degree.

If ACBO is a rectangle, then opposite sides are equal. Also as two tangents are equal and both the radius are equal, its a square.

Triangle AOB is a 45-45-90 triangle and hence k= root 2 and AO=OB=BC=AC= 1

Applying the theorem:

r^2+r^2= k^2
2r^2= k^2

r= k/root 2
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Re: The circle with center C shown above is tangent to both axes  [#permalink]

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New post 05 Sep 2017, 17:21
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 have a right triangle with sides r, r, k. It is an isosceles.

\(k^2 = r^2 + r^2\)
\(k^2 = 2 * r^2\)
\(k = \sqrt{2}*r\)
\(r = k / \sqrt{2}\)

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

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New post 05 Sep 2017, 17:39
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) \(\frac{k}{3}\)


\((\sqrt{x}+\sqrt{y}) / \sqrt{x + y}\)
\(= (\sqrt{4} + \sqrt{6}) / \sqrt{10}\)
\(= ( 2 + \sqrt{6} ) \sqrt{10} / \sqrt{10} \sqrt{10}\)
\(= ( 2\sqrt{10} + \sqrt{60} ) / 10\)
\(= (2\sqrt{10} + 2\sqrt{15}) / 10\)
\(= ( \sqrt{10} + \sqrt{15} ) / 5\)

The answer is D.
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Re: The circle with center C shown above is tangent to both axes   [#permalink] 05 Sep 2017, 17:39

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