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Manager  S
Joined: 14 Sep 2015
Posts: 64
Location: India
GMAT 1: 700 Q45 V40 GPA: 3.41
In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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13 00:00

Difficulty:   75% (hard)

Question Stats: 67% (03:18) correct 33% (02:54) wrong based on 86 sessions

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In a rectangular coordinate plane, AB is the diameter of a circle and point C lies on the circle. If the coordinates of points A and B are (-1,0) and (5,0), and the area of triangle ABC is $$6\sqrt{2}$$square units, which of the following can be the coordinates of point C?

A. (0, 2$$\sqrt{2}$$)
B. (1, 2$$\sqrt{2}$$)
C. ($$\sqrt{2}$$,2)
D. (2,$$\sqrt{2}$$)
E. (2$$\sqrt{2}$$,1)
Retired Moderator V
Status: Long way to go!
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In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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niteshwaghray wrote:
In a rectangular coordinate plane, AB is the diameter of a circle and point C lies on the circle. If the coordinates of points A and B are (-1,0) and (5,0), and the area of triangle ABC is $$6\sqrt{2}$$square units, which of the following can be the coordinates of point C?

A. (0, 2$$\sqrt{2}$$)
B. (1, 2$$\sqrt{2}$$)
C. ($$\sqrt{2}$$,2)
D. (2,$$\sqrt{2}$$)
E. (2$$\sqrt{2}$$,1)

Attachment: Capture.PNG [ 12.32 KiB | Viewed 1766 times ]

We have $$AB=6 \implies R = 3$$.

$$I$$ is the center of that circle, then we have $$I(2,0)$$

The equation of that circle is $$(I): \; \; (x-2)^2 + y^2 = 3^2$$

The coordinates of $$C(x_C, y_C)$$. We have $$(x_C-2)^2 + y_C^2 = 9$$

Also, we have
$$S_{ABC}=6\sqrt{2} \implies \frac{AB \times |y_C|}{2}=6\sqrt{2} \\ \implies |y_C|=2\sqrt{2} \implies y_C^2=8$$
$$\implies (x_C-2)^2 = 1 \implies x_C = 3$$ or $$x_C = 1$$.

Only choice B fits the roots. B is the correct answer.
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Intern  B
Joined: 03 Sep 2018
Posts: 7
In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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2
Analysis (51 seconds): Draw out a rough estimation of the coordinate plane, mark the points and draw a circle. Notice that I can use the area of a triangle to get a value for Y for point C.

Strategy: Find Y, Eliminate

Find Y (35 seconds)
$$Ta = \frac{1}{2}*b*h$$

$$\frac{1}{2}*6*h = 6\sqrt{2}$$

$$3*h = 6\sqrt{2}$$

$$Y = 6\sqrt{2} / 3$$

$$Y = 2\sqrt{2}$$

Eliminate (20 seconds)
• A: With only two choices left, logically it can only be the one closest to the centre of the circle so eliminate A.
• C: Wrong Y
• D: Wrong Y
• E: Wrong Y

Time: 1:46
Intern  B
Joined: 24 Jun 2017
Posts: 30
Re: In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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jameslewis wrote:
Analysis (51 seconds): Draw out a rough estimation of the coordinate plane, mark the points and draw a circle. Notice that I can use the area of a triangle to get a value for Y for point C.

Strategy: Find Y, Eliminate

Find Y (35 seconds)
$$Ta = \frac{1}{2}*b*h$$

$$\frac{1}{2}*6*h = 6\sqrt{2}$$

$$3*h = 6\sqrt{2}$$

$$Y = 6\sqrt{2} / 3$$

$$Y = 2\sqrt{2}$$

Eliminate (20 seconds)
• A: With only two choices left, logically it can only be the one closest to the centre of the circle so eliminate A.
• C: Wrong Y
• D: Wrong Y
• E: Wrong Y

Time: 1:46

though you are getting the correct answer , but i guess the method is wrong as AB is the diameter of the circle and point C on the circle so <C =90 deg so , AB will be the hypotenuse not the base of the triangle , as you have used it as base of 6 units to eliminate answer choices
let me know if i am wrong Manager  S
Joined: 26 Mar 2019
Posts: 107
Concentration: Finance, Strategy
Re: In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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broall wrote:
niteshwaghray wrote:
In a rectangular coordinate plane, AB is the diameter of a circle and point C lies on the circle. If the coordinates of points A and B are (-1,0) and (5,0), and the area of triangle ABC is $$6\sqrt{2}$$square units, which of the following can be the coordinates of point C?

A. (0, 2$$\sqrt{2}$$)
B. (1, 2$$\sqrt{2}$$)
C. ($$\sqrt{2}$$,2)
D. (2,$$\sqrt{2}$$)
E. (2$$\sqrt{2}$$,1)

Attachment:
Capture.PNG

We have $$AB=6 \implies R = 3$$.

$$I$$ is the center of that circle, then we have $$I(2,0)$$

The equation of that circle is $$(I): \; \; (x-2)^2 + y^2 = 3^2$$

The coordinates of $$C(x_C, y_C)$$. We have $$(x_C-2)^2 + y_C^2 = 9$$

Also, we have
$$S_{ABC}=6\sqrt{2} \implies \frac{AB \times |y_C|}{2}=6\sqrt{2} \\ \implies |y_C|=2\sqrt{2} \implies y_C^2=8$$
$$\implies (x_C-2)^2 = 1 \implies x_C = 3$$ or $$x_C = 1$$.

Only choice B fits the roots. B is the correct answer.

Hi, broall , I did not get the equation of the circle from your explanation. Would you be so kind to explain how you derive it? Thanks.
Manager  B
Joined: 19 Feb 2019
Posts: 101
Concentration: Marketing, Statistics
Re: In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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Quote:
Eliminate (20 seconds)
A: With only two choices left, logically it can only be the one closest to the centre of the circle so eliminate A.
C: Wrong Y
D: Wrong Y
E: Wrong Y

How did you eliminate option A??
How did you arrive at x co ordinate as 1? it could have also been 0 and y co ordinate as 2 root 2

Can you pls explain
Manager  G
Joined: 28 Feb 2014
Posts: 209
Location: India
GPA: 3.97
WE: Engineering (Education)
Re: In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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niteshwaghray wrote:
In a rectangular coordinate plane, AB is the diameter of a circle and point C lies on the circle. If the coordinates of points A and B are (-1,0) and (5,0), and the area of triangle ABC is $$6\sqrt{2}$$square units, which of the following can be the coordinates of point C?

A. (0, 2$$\sqrt{2}$$)
B. (1, 2$$\sqrt{2}$$)
C. ($$\sqrt{2}$$,2)
D. (2,$$\sqrt{2}$$)
E. (2$$\sqrt{2}$$,1)

AB = 6 units
When AB is the base of the triangle and h is height of the triangle

Area = (1/2) * AB * h = 6\sqrt{2}
h = 2\sqrt{2}

Options C, D and E are eliminated

option A not possible as it is lying on the x axis

B is correct.
Intern  B
Joined: 03 Sep 2018
Posts: 1
Re: In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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kapil, A is lying on the y axis.
I still don't understand why eliminate A

Posted from my mobile device Re: In a rectangular coordinate plane, AB is the diameter of a circle and   [#permalink] 09 Dec 2019, 03:49
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