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# In a rectangular coordinate plane, AB is the diameter of a circle and

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Manager
Joined: 14 Sep 2015
Posts: 65
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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31 May 2017, 23:15
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Question Stats:

72% (02:25) correct 28% (02:03) wrong based on 89 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)
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In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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01 Jun 2017, 00:40
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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 1099 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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In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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25 Sep 2018, 10:05
1
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
In a rectangular coordinate plane, AB is the diameter of a circle and &nbs [#permalink] 25 Sep 2018, 10:05
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