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In the figure above, O is the center of the circle. Line AB intersects

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In the figure above, O is the center of the circle. Line AB intersects [#permalink]

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New post 23 Nov 2017, 23:19
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In the figure above, O is the center of the circle. Line AB intersects the circle only at point B, and line DC intersects the circle only at point C. If the circle has radius of 2, then AC =

(A) 4
(B) 2√2
(C) 4 + √2
(D) 4 + √3
(E) 2 + 2√2


[Reveal] Spoiler:
Attachment:
2017-11-23_2025_001.png
2017-11-23_2025_001.png [ 8.65 KiB | Viewed 443 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, O is the center of the circle. Line AB intersects [#permalink]

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New post 24 Nov 2017, 01:27
E.

Produce OB and it shall make a right angle at point of contact.
Triangle ABO will be right and isosceles.

AO can be found from pythagoras theorem.

AO+radius is the answer.


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Re: In the figure above, O is the center of the circle. Line AB intersects [#permalink]

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New post 24 Nov 2017, 02:16
Bunuel wrote:
Image
In the figure above, O is the center of the circle. Line AB intersects the circle only at point B, and line DC intersects the circle only at point C. If the circle has radius of 2, then AC =

(A) 4
(B) 2√2
(C) 4 + √2
(D) 4 + √3
(E) 2 + 2√2


[Reveal] Spoiler:
Attachment:
2017-11-23_2025_001.png


OB is the radius of the circle and AB is the tangent. Radius is perpendicular to tangent at the point of intersection. In triangle OBA, angle A is 45 degrees, angle ABO is 90 degrees and hence angle BOA is also 45 degrees. This is a 45-45-90 triangle in which ratio of the sides will be \(1:1:\sqrt{2}\)
Using pythagorean theorem, AO \(= 2\sqrt{2}\)

AC = AO + OC = \(2\sqrt{2} + 2\)

Answer (E)
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In the figure above, O is the center of the circle. Line AB intersects [#permalink]

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New post 25 Nov 2017, 10:25
Bunuel wrote:
Image
In the figure above, O is the center of the circle. Line AB intersects the circle only at point B, and line DC intersects the circle only at point C. If the circle has radius of 2, then AC =

(A) 4
(B) 2√2
(C) 4 + √2
(D) 4 + √3
(E) 2 + 2√2


[Reveal] Spoiler:
Attachment:
The attachment 2017-11-23_2025_001.png is no longer available


ABO is a right isosceles triangle, hence \(BO = AB = r = 2\), hence the side AO = \(2\sqrt{2}\). \(AO + r = 2\sqrt{2} + 2\). The answer is E
Attachments

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Re: In the figure above, O is the center of the circle. Line AB intersects [#permalink]

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New post 25 Nov 2017, 11:04
Bunuel wrote:
Image
In the figure above, O is the center of the circle. Line AB intersects the circle only at point B, and line DC intersects the circle only at point C. If the circle has radius of 2, then AC =

(A) 4
(B) 2√2
(C) 4 + √2
(D) 4 + √3
(E) 2 + 2√2

[Reveal] Spoiler:
Attachment:
The attachment 2017-11-23_2025_001.png is no longer available

Attachment:
ccccc.png
ccccc.png [ 16.4 KiB | Viewed 146 times ]

The length of AC = (length of hypotenuse, AO, of an isosceles right triangle) + (radius, CO)

Find length of AO

Connect O with B
Point B is tangent, so the radius is perpendicular to line AB and creates a right angle at point B

The triangle hence is a 45-45-90 isosceles right triangle, with side lengths in ratio*
\(x: x: x\sqrt{2}\)

One leg, BO \(= x\) = radius = \(2\)
Other leg, AB = BO = \(x = r = 2\)
Hypotenuse, AO = \(x\sqrt{2}=\)\(2\sqrt{2}\)

Add segment CO to get length of AC

The rest of the length of AC is
CO = r = \(2\)

Length of AC = \(2\sqrt{2} + 2\)

Answer E

*OR Pythagorean theorem:
\(leg^2 + leg^2 = hypotenuse, h^2\)
\(2^2 + 2^2 = h^2\)
\(8 = h^2\)
\(\sqrt{4*2} = \sqrt{h^2}\)
\(h = 2\sqrt{2}\) = length of AO

Kudos [?]: 393 [0], given: 640

Re: In the figure above, O is the center of the circle. Line AB intersects   [#permalink] 25 Nov 2017, 11:04
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In the figure above, O is the center of the circle. Line AB intersects

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