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Re: Circle O passes through points (9, 2√3), (7, 0), and (11, 0) [#permalink]
4
The diameter is the largest distance between two points in a circle,
If the diameter of the circle is 4 all the given points in the question cannot fall on the circle.
Also from the points given (7,0) and (4,0) gives us a chord and the chord is always smaller than the diameter

SO, the answer is A
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Circle O passes through points (9, 2√3), (7, 0), and (11, 0) [#permalink]
1
IMO analytical solution for distance between all points given
distance between points \((9, 2\sqrt{3})\) and \((11, 0)\) results in \(\sqrt{21}\), which is a potential diameter or the maximum distance between two points on the circle
since \(\sqrt{21}\) > \(4\), select answer A
KarunMendiratta wrote:
Circle O passes through points \((9, 2\sqrt{3})\), \((7, 0)\) and \((11, 0)\)

Quantity A
Quantity B
Diameter of the circle
4


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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Circle O passes through points (9, 2√3), (7, 0), and (11, 0) [#permalink]
motion2020 wrote:
IMO analytical solution for distance between all points given
distance between points \((9, 2\sqrt{3})\) and \((11, 0)\) results in \(\sqrt{21}\), which is a potential diameter or the maximum distance between two points on the circle
since \(\sqrt{21}\) > \(4\), select answer A
KarunMendiratta wrote:
Circle O passes through points \((9, 2\sqrt{3})\), \((7, 0)\) and \((11, 0)\)

Quantity A
Quantity B
Diameter of the circle
4


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Can you show the calculation of how did you get \(\sqrt{21}\) because I am still getting \(\sqrt{16}\) which is 4?
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Re: Circle O passes through points (9, 2√3), (7, 0), and (11, 0) [#permalink]
1
sorry, I have miscalculated
it looks like there are three equal sides now, and we got equilateral triangle circumsribed by circle
then considering the center of equialteral triangle and 30-60-90 tringle with base = side/2 we can find that radius (hypothenuse of 30-60-90 triangle) such as 2*side/(2*sq.root(3)), radius=4/sq.root(3) and diameter=8/sq.root(3). Since sq.root(3)<2, we conclude that diameter > 4 and answer is A.

koala wrote:
motion2020 wrote:
IMO analytical solution for distance between all points given
distance between points \((9, 2\sqrt{3})\) and \((11, 0)\) results in \(\sqrt{21}\), which is a potential diameter or the maximum distance between two points on the circle
since \(\sqrt{21}\) > \(4\), select answer A
KarunMendiratta wrote:
Circle O passes through points \((9, 2\sqrt{3})\), \((7, 0)\) and \((11, 0)\)

Quantity A
Quantity B
Diameter of the circle
4


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Can you show the calculation of how did you get \(\sqrt{21}\) because I am still getting \(\sqrt{16}\) which is 4?

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Re: Circle O passes through points (9, 2√3), (7, 0), and (11, 0) [#permalink]
1
I did this on another way,
just draw a circle with all the points on the plane. If you look thoroughly, you will notice that the point (9,2√3) is the center of the circle.
And rest of the two points lie on the circumference.
So lets find the distance between center and the any point in the circumference. r= (2√3-0)^2 + (9-7)^2 = 16. now take square root of 16 , which is 4 (our radius)
Now make it diameter 2*4=8 , hence QA is bigger. Boom
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Re: Circle O passes through points (9, 2√3), (7, 0), and (11, 0) [#permalink]
KarunMendiratta wrote:
Explanation:

Let the centre be \((h, k)\) which will be equidistant from all 3 points.

\(Radius^2 = (11 - h)^2 + (0 - k)^2 = (7 - h)^2 + (0 - k)^2 = (9 - h)^2 + (2\sqrt{3} - k)^2\)

Using \((11 - h)^2 + (0 - k)^2 = (7 - h)^2 + (0 - k)^2\)
\(121 + h^2 - 22h + k^2= 49 + h^2 - 14h + k^2\)
\(121 - 49 = 8h\)
\(72 = 8h\)
\(h = 9\)

Plugging in the value of \(h = 9\) in \((7 - h)^2 + (0 - k)^2 = (9 - h)^2 + (2\sqrt{3} - k)^2\);

\((7 - 9)^2 + (0 - k)^2 = (9 - 9)^2 + (2\sqrt{3} - k)^2\)
\(4 + k^2 = 0 + 12 + k^2 - 4\sqrt{3}k\)
\(4\sqrt{3}k = 8\)
\(\sqrt{3}k = 2\)
\(k = \frac{2}{\sqrt{3}}\)

So, the centre is \((9, \frac{2}{\sqrt{3}})\)

\(Radius^2 = (9 - 7)^2 + (\frac{2}{\sqrt{3}} - 0)^2 = 4 + \frac{4}{3} = \frac{16}{3}\)

\(Radius = \frac{4}{\sqrt{3}}\)
Diameter = \(2(\frac{4}{\sqrt{3}}) = \frac{8}{\sqrt{3}}\)

Col. A: \(\frac{8}{\sqrt{3}}\)
Col. B: \(4\)

Dividing by \(4\) and multiplying by \(\sqrt{3}\) on both sides;

Col. A: \(2\)
Col. B: \(\sqrt{3}\)

Hence, option A

great explanation .
any easy method than this to sole in less time .
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Re: Circle O passes through points (9, 23), (7, 0), and (11, 0) [#permalink]
1
I have a much simpler solution.
If 4 was a diameter, which implies that (9,0) is circle's center, then point (9,2√3) cannot be on the circumference. Hence, our assumed diameter is just a chord, therefore actual diameter must be bigger. Thus, answer is A
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Re: Circle O passes through points (9, 23), (7, 0), and (11, 0) [#permalink]
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Re: Circle O passes through points (9, 23), (7, 0), and (11, 0) [#permalink]
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