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

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

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New post 01 Jun 2017, 00:15
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Question Stats:

66% (03:28) correct 34% (02:35) wrong based on 68 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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New post 01 Jun 2017, 01:40
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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)


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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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New post 25 Sep 2018, 11:05
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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.
  • B: Correct Answer
  • C: Wrong Y
  • D: Wrong Y
  • E: Wrong Y

Answer: B
Time: 1:46
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Re: In a rectangular coordinate plane, AB is the diameter of a circle and  [#permalink]

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New post 17 Mar 2019, 13:34
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.
  • B: Correct Answer
  • C: Wrong Y
  • D: Wrong Y
  • E: Wrong Y

Answer: B
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 :)
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Re: In a rectangular coordinate plane, AB is the diameter of a circle and   [#permalink] 17 Mar 2019, 13:34
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In a rectangular coordinate plane, AB is the diameter of a circle and

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