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niteshwaghray
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In a rectangular coordinate plane, AB is the diameter of a circle and 01 Jun 2017, 00:15
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

63% (03:05) correct 37% (02:59) wrong based on 167 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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B
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In a rectangular coordinate plane, AB is the diameter of a circle and 01 Jun 2017, 01:40
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niteshwaghray
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
Capture.PNG [ 12.32 KiB | Viewed 5315 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 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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In a rectangular coordinate plane, AB is the diameter of a circle and 17 Mar 2019, 13:34
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jameslewis
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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In a rectangular coordinate plane, AB is the diameter of a circle and 24 Nov 2019, 11:36
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niteshwaghray
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.
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In a rectangular coordinate plane, AB is the diameter of a circle and 08 Dec 2019, 23:11
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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.
B: Correct Answer
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
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In a rectangular coordinate plane, AB is the diameter of a circle and 08 Dec 2019, 23:57
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niteshwaghray
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.
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In a rectangular coordinate plane, AB is the diameter of a circle and 09 Dec 2019, 03:49
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kapil, A is lying on the y axis.
I still don't understand why eliminate A

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In a rectangular coordinate plane, AB is the diameter of a circle and 19 Mar 2021, 22:33
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Formula for Circle:

(X - a)^2 + (Y - b)^2 = (r)^2

Where point (a , b) is the Center of the Circle and r is the Radius


Since we are given the 2 points as the Diameter of the Circle, the Diameter = D = 6 units along X axis from point A to point D

The Radius = r = 3

The Center will be the mid point of the Diameter on the X Axis, which occurs at point (2 , 0)

The formula of the Circle is given by:

(X - 2)^2 + (Y)^2 = 9

Now, since we are told that point C is on the circumference of the circle, we have an inscribed 90 degree triangle with the hypotenuse as the diameter.


Using the Hypotenuse as the Base of the triangle, we can find the Height of this Triangle that extends from the Base (Diameter/Hypotenuse) to the 90 degree vertex at C


Area of triangle = (1/2) * (6) * (h) = (6) * sqrt(2)

Perpendicular Height from Diameter/Hypotenuse to the vertex at Point C = h = 2 * sqr(2)


Given that the Base we used is on the X Axis, this perpendicular height - h - will be given by the Y Coordinate of Point C.

In other words, the height will be equal to the distance as measured by the Y Axis from Y = 0 to vertex C ——> which is given by the Y coordinate of point C

Thus, the Y coordinate of point C must either be:

2 * sqrt(2)

or

(-)2 * sqrt(2)


We can eliminate 3 options, leaving only A and B.

If you are short on time, you can visualize the circle on the graph paper and realize that the Point given by (0 , 2*sqrt(2) )

Will be Outside the edge of the Circle.


To confirm that answer B lies on the circle, you can plug the coordinates into the equation for the circle that we have.


When you plug the coordinates given in answer B into:

(X - 2)^2 + (Y)^2 = 9

You will find that the equation is satisfied and that vertex C lies on the circumference of the circle.

Answer (B)


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

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In a rectangular coordinate plane, AB is the diameter of a circle and 08 May 2022, 19:12
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