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WE:Education (Education)
Circle O passes through points (9, 2√3), (7, 0), and (11, 0)
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13 Aug 2021, 08:21
13 Aug 2021, 08:21
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Question Stats:
37% (02:06) correct
62% (01:25) wrong based on 51 sessions
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Circle O passes through points \((9, 2\sqrt{3})\), \((7, 0)\) and \((11, 0)\)
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
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
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Circle O passes through points (9, 2√3), (7, 0), and (11, 0)
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15 Aug 2021, 06:50
15 Aug 2021, 06:50
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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
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
General Discussion
Re: Circle O passes through points (9, 2√3), (7, 0), and (11, 0)
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13 Aug 2021, 11:10
13 Aug 2021, 11:10
my stand is c.
