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Given two intersecting circles..
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Updated on: 20 Jan 2019, 11:50
Updated on: 20 Jan 2019, 11:50
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Given two intersecting circles, one with radius 15 and another with radius 25. If the distance between the two intersecting points is 20, then find the distance between the centers of the two circles?
Show: :: OA
34.09
Originally posted by Sonalika42 on 28 Aug 2016, 04:55.
Last edited by Carcass on 20 Jan 2019, 11:50, edited 1 time in total.
Last edited by Carcass on 20 Jan 2019, 11:50, edited 1 time in total.
Edited by Carcass
Retired Moderator
Joined: 07 Jun 2014
Posts: 4806
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Re: Given two intersecting circles..
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01 Oct 2018, 13:22
01 Oct 2018, 13:22
4
Expert Reply
In this case let the circles intersect at B and F respectively, as shown in figure below:
Inkedintersecting.gif [ 17.93 KiB | Viewed 5845 times ]
Now line AC is a perpedicular bisector of BF at point D and AB=15 and BC=25. Hence to find the distance we need to calculate AD and DC.
Now triangle ABD is right triangle AD=\(\sqrt{AB^2-BD^2}=\sqrt{15^2-10^2}=11.18\)
Now triangle CBD is right triangle CD=\(\sqrt{BC^2-BD^2}=\sqrt{25^2-10^2}=22.91\).
Hence total distance AC = 11.18+22.91=34.09.
The portion marked in red are properties of a circle.
_________________
Attachment:
Inkedintersecting.gif [ 17.93 KiB | Viewed 5845 times ]
Now line AC is a perpedicular bisector of BF at point D and AB=15 and BC=25. Hence to find the distance we need to calculate AD and DC.
Now triangle ABD is right triangle AD=\(\sqrt{AB^2-BD^2}=\sqrt{15^2-10^2}=11.18\)
Now triangle CBD is right triangle CD=\(\sqrt{BC^2-BD^2}=\sqrt{25^2-10^2}=22.91\).
Hence total distance AC = 11.18+22.91=34.09.
The portion marked in red are properties of a circle.
_________________
Sandy
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Re: Given two intersecting circles..
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31 Oct 2018, 05:53
31 Oct 2018, 05:53
The answer is 20. The distance between intersection points is more than a circle's radius, hence one of the circle is pretty smaller than the other.
Re: Given two intersecting circles..
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02 Nov 2018, 09:14
02 Nov 2018, 09:14
One circle has to be larger than the other one for the distance to be greater than one of the radii
Re: Given two intersecting circles..
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03 Dec 2018, 09:11
03 Dec 2018, 09:11
Sandy, how are we supposed to approximate square roots?
Re: Given two intersecting circles..
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17 Dec 2018, 08:15
17 Dec 2018, 08:15
sandy wrote:
In this case let the circles intersect at B and F respectively, as shown in figure below:
Now line AC is a perpedicular bisector of BF at point D and AB=15 and BC=25. Hence to find the distance we need to calculate AD and DC.
Now triangle ABD is right triangle AD=\(\sqrt{AB^2-BD^2}=\sqrt{15^2-10^2}=11.18\)
Now triangle CBD is right triangle CD=\(\sqrt{BC^2-BD^2}=\sqrt{25^2-10^2}=22.91\).
Hence total distance AC = 11.18+22.91=34.09.
The portion marked in red are properties of a circle.
Attachment:
Inkedintersecting.gif
Now line AC is a perpedicular bisector of BF at point D and AB=15 and BC=25. Hence to find the distance we need to calculate AD and DC.
Now triangle ABD is right triangle AD=\(\sqrt{AB^2-BD^2}=\sqrt{15^2-10^2}=11.18\)
Now triangle CBD is right triangle CD=\(\sqrt{BC^2-BD^2}=\sqrt{25^2-10^2}=22.91\).
Hence total distance AC = 11.18+22.91=34.09.
The portion marked in red are properties of a circle.
I can't tell where you got the 10 from, it shouldn't be as simple as just subtracting 25 from 15 that doesn't make sense.
Re: Given two intersecting circles..
[#permalink]
20 Jan 2019, 07:07
20 Jan 2019, 07:07
QuantumWonder wrote:
sandy wrote:
In this case let the circles intersect at B and F respectively, as shown in figure below:
Now line AC is a perpedicular bisector of BF at point D and AB=15 and BC=25. Hence to find the distance we need to calculate AD and DC.
Now triangle ABD is right triangle AD=\(\sqrt{AB^2-BD^2}=\sqrt{15^2-10^2}=11.18\)
Now triangle CBD is right triangle CD=\(\sqrt{BC^2-BD^2}=\sqrt{25^2-10^2}=22.91\).
Hence total distance AC = 11.18+22.91=34.09.
The portion marked in red are properties of a circle.
Attachment:
Inkedintersecting.gif
Now line AC is a perpedicular bisector of BF at point D and AB=15 and BC=25. Hence to find the distance we need to calculate AD and DC.
Now triangle ABD is right triangle AD=\(\sqrt{AB^2-BD^2}=\sqrt{15^2-10^2}=11.18\)
Now triangle CBD is right triangle CD=\(\sqrt{BC^2-BD^2}=\sqrt{25^2-10^2}=22.91\).
Hence total distance AC = 11.18+22.91=34.09.
The portion marked in red are properties of a circle.
I can't tell where you got the 10 from, it shouldn't be as simple as just subtracting 25 from 15 that doesn't make sense.
It is in the question,
Given two intersecting circles, one with radius 15 and another with radius 25. If the distance between the two intersecting points is 20, then find the distance between the centers of the two circles?
Then perpendicular bisector properties of a circle.
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Given two intersecting circles..
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25 Feb 2021, 08:21
25 Feb 2021, 08:21
1
Incognito wrote:
Sandy, how are we supposed to approximate square roots?
You are not required to, you will have an onscreen calculator in the exam!
Howeover, you can remember these 5 values in particular, that will help you:
\(\sqrt{2} = 1.41, \sqrt{3} = 1.73, \sqrt{5} = 2.23, \sqrt{7} = 2.64, \sqrt{10} = 3.16\)
_________________
I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, India
https://www.instagram.com/prepster_education/
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, India
https://www.instagram.com/prepster_education/
Re: Given two intersecting circles..
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22 Jul 2023, 11:01
22 Jul 2023, 11:01
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