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# The circle shown is tangent to both the x and y-axes. If the length of

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Math Expert
Joined: 02 Sep 2009
Posts: 52390
The circle shown is tangent to both the x and y-axes. If the length of  [#permalink]

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01 Apr 2018, 09:39
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Question Stats:

93% (01:25) correct 7% (01:16) wrong based on 72 sessions

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The circle shown is tangent to both the x and y-axes. If the length of the segment from the circle’s center C to the origin, O, is 6, what is the circle’s radius?

(A) 6
(B) $$3 \sqrt{2}$$
(C) $$2 \sqrt{3}$$
(D) 3
(E) 2

Attachment:

2018-04-01_2137.png [ 11.17 KiB | Viewed 848 times ]

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Concentration: General Management, Marketing
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The circle shown is tangent to both the x and y-axes. If the length of  [#permalink]

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01 Apr 2018, 09:48
Bunuel wrote:

The circle shown is tangent to both the x and y-axes. If the length of the segment from the circle’s center C to the origin, O, is 6, what is the circle’s radius?

(A) 6
(B) $$3 \sqrt{2}$$
(C) $$2 \sqrt{3}$$
(D) 3
(E) 2

Attachment:
2018-04-01_2137.png

let the center be (x,y)
From diag it is clear (x,y) is (r,r)

so, by pythagorus theorem (\sqrt{2})*r=6
r=3*\sqrt{2}
option B
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Joined: 03 Nov 2017
Posts: 14
GMAT 1: 610 Q47 V28
Re: The circle shown is tangent to both the x and y-axes. If the length of  [#permalink]

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01 Apr 2018, 19:47
IMO : B

Given x and y axis are tangent to the circle, they make a 90-degree angle when radius of the circle intersects with them.

This makes the two radius + the axis into a square and from there we can determine the side by following 45 45 90 rule
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Joined: 19 Oct 2017
Posts: 4
The circle shown is tangent to both the x and y-axes. If the length of  [#permalink]

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03 Apr 2018, 16:21
Since the Circle is tangent to the Y and X axis, we can draw a line from C to the Y axis and the X axis.
We also know that those lines form 90 degrees at the axis.

Method 1:
45-45-90 triangle
Ratio: 1:1:√2 which we can convert to x: x: 6. Basically we need a number, when multiplied by √2 will give us 6. 3√2 is that value.

Method 2:
After we draw those two lines to the X/Y axis, we can determine that those two lines, and with reference point O, will form a square.
Thus the length and width must be the same. The diagonal of a square is always Length of a side * √2
Again, 3√2 is that value.
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Re: The circle shown is tangent to both the x and y-axes. If the length of  [#permalink]

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05 Apr 2018, 10:53
You can draw a square from the circle center, where the legs run to the X axis and y axis. The line running from the origin to the center of the circle is the diameter of the square.

Area of square: Area = S^2 or Area = (D^2)/ 2
Area = (6^2)/2 = 18

The legs drawn to form the square are also the radius of ther circle. The square root of 18 will equal the radius.
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Re: The circle shown is tangent to both the x and y-axes. If the length of  [#permalink]

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06 Apr 2018, 07:51
Bunuel wrote:

The circle shown is tangent to both the x and y-axes. If the length of the segment from the circle’s center C to the origin, O, is 6, what is the circle’s radius?

(A) 6
(B) $$3 \sqrt{2}$$
(C) $$2 \sqrt{3}$$
(D) 3
(E) 2

Attachment:
2018-04-01_2137.png

We can drop a perpendicular from point C to the x-axis, and call the point of intersection D so that triangle OCD is a 45-45-90 triangle (note: CD is a radius of the circle that is perpendicular to OD which is lying on the x-axis). We will use the fact that the ratio of a side to the hypotenuse of a 45-45-90 triangle is x : x√2.

If we let CD = OD = x = the length of (either) side of the 45-45-90 triangle, then the length of the hypotenuse of the 45-45-90 triangle is x√2. From the diagram, we know that the length of the hypotenuse is 6. Thus, we can create the equation:

x√2 = 6

x = 6/√2

We need to rationalize the denominator. Multiplying by √2/√2, we have:

x = 6√2/2 = 3√2

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Joined: 06 Dec 2017
Posts: 7
Re: The circle shown is tangent to both the x and y-axes. If the length of  [#permalink]

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09 Apr 2018, 01:38
Hello, I don't understand the need to rationalize 6√2 to 3√2. can someone please elaborate?
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Joined: 14 Oct 2017
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The circle shown is tangent to both the x and y-axes. If the length of  [#permalink]

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15 Apr 2018, 02:32
1
antarcticsugar wrote:
Hello, I don't understand the need to rationalize 6√2 to 3√2. can someone please elaborate?

We have $$6/\sqrt{2}$$ that we multiply by $$\sqrt{2}/\sqrt{2}$$.

This results in $$(6*\sqrt{2})/2$$, which can be simplified to $$3\sqrt{2}$$.

I hope that helps
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The circle shown is tangent to both the x and y-axes. If the length of &nbs [#permalink] 15 Apr 2018, 02:32
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# The circle shown is tangent to both the x and y-axes. If the length of

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