In the coordinate plane, a circle with its center on the negative x-ax

Question

In the coordinate plane, a circle with its center on the negative x-axis has a radius of 12 units, and passes through (0, 6) and (0, – 6). What is the area of the part of this circle in the first quadrant?
(A) \(6\pi - 12\sqrt{3}\)
(B) \(12\pi - 12\sqrt{3}\)
(C) \(12\pi - 18\sqrt{3}\)
(D) \(24\pi - 18\sqrt{3}\)
(E) \(24\pi - 36\sqrt{3}\)

This is a question from a set of 15 challenging GMAT Quant practice questions. For the whole collection, as well as the OE with diagram for this question, see:
Challenging GMAT Math Practice Questions

Mike

saswata4s · Accepted Answer

The circle passes through (0, 6) and (0, – 6). Therefore, if we draw a chord between these two points, the length of the chord will be 12.

Again, the radius is 12. So if we connect the center and (0, 6) and (0, – 6), we get an equilateral triangle which has sides of length 12. Hence the angle created at the center by the chord is 60.
So, the area OAB = π(12)^2 * 60/360 = π*12*12*(1/6) = π*12*2 = 24π.

Now, the area of the triangle = (\sqrt{3} /4)*12*12 = \sqrt{3}*3*12 = 36\sqrt{3}.

So, the area of the of the part of this circle in the first quadrant = 1/2(24π - 36\sqrt{3}) = (12π - 18\sqrt{3})

Cheers.

mikemcgarry · Answer

Dear saswata4s , Thank you very much for finding this typo. :-)

You are 100% correct. I corrected the answer choices above. Thanks again, Mike McGarry :-)

kakakakaak · Answer

in last wy are we myltiplyin by 1/2

Samuel0709 · Answer

Question is asking us for area in first quadrant only. Since area enclosed by in first quandrant = area enclosed in second quadrant we are dividing the total area by 2, hope this is clear

KarishmaB · Answer

We need the area OPMO. People with some trigonometry knowledge can solve this fairly quickly by recognising that Sin of angle PCO is 1/2 which means that angle PCO is 30 degrees. Screenshot 2023-03-25 at 12.45.48 PM.png Area OPMO = Area of sector CPMC - Area of triangle CPO = \frac{30}{360}* \pi * 12^2 - \frac{1}{2}*6*\sqrt{12^2-6^2} Area OPMO = 12\pi - 18\sqrt{3} Answer (C) In case one doesn't want to use trigonometry, one could recognize that triangle CPQ is an equilateral triangle. So angle PCQ is 60 degrees which means angle PCO is 30 degrees and then proceed as before.

bumpbot · Answer

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In the coordinate plane, a circle with its center on the negative x-ax

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In the coordinate plane, a circle with its center on the negative x-ax

Prep Toolkit

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