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

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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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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


[Reveal] Spoiler:
Attachment:
2018-04-01_2137.png
2018-04-01_2137.png [ 11.17 KiB | Viewed 590 times ]
[Reveal] Spoiler: OA

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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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New post 01 Apr 2018, 10:48
Bunuel wrote:
Image

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


[Reveal] Spoiler:
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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Re: The circle shown is tangent to both the x and y-axes. If the length of [#permalink]

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New post 01 Apr 2018, 20:47
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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The circle shown is tangent to both the x and y-axes. If the length of [#permalink]

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New post 03 Apr 2018, 17:21
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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New post 05 Apr 2018, 11:53
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

Answer B
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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New post 06 Apr 2018, 08:51
Bunuel wrote:
Image

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

[Reveal] Spoiler:
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

Answer: B
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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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New post 09 Apr 2018, 02:38
Hello, I don't understand the need to rationalize 6√2 to 3√2. can someone please elaborate?
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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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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   [#permalink] 15 Apr 2018, 03:32
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