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

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Math Expert
Joined: 02 Sep 2009
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In the figure above, O is the center of the circle. Line AB intersects [#permalink]

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23 Nov 2017, 23:19
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75% (01:08) correct 25% (01:04) wrong based on 28 sessions

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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 [ 8.65 KiB | Viewed 443 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 135185 [1], given: 12671

Intern
Joined: 29 Aug 2016
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Re: In the figure above, O is the center of the circle. Line AB intersects [#permalink]

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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.

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7786

Kudos [?]: 18094 [0], given: 236

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

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24 Nov 2017, 02:16
Bunuel wrote:

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 = $$2\sqrt{2} + 2$$

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Kudos [?]: 18094 [0], given: 236

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

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25 Nov 2017, 10:25
Bunuel wrote:

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

circle.jpg [ 20.66 KiB | Viewed 163 times ]

Kudos [?]: 2 [0], given: 57

VP
Joined: 22 May 2016
Posts: 1105

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

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25 Nov 2017, 11:04
Bunuel wrote:

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 [ 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 = $$2\sqrt{2} + 2$$

*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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