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Given two intersecting circles..

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Sonalika42
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Given two intersecting circles.. [#permalink] New post   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?

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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.
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Re: Given two intersecting circles.. [#permalink] New post   01 Oct 2018, 13:22
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In this case let the circles intersect at B and F respectively, as shown in figure below:

Attachment:
Inkedintersecting.gif
Inkedintersecting.gif [ 17.93 KiB | Viewed 6033 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.
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Re: Given two intersecting circles.. [#permalink] New post   29 Sep 2018, 23:32
what is the answer??
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Re: Given two intersecting circles.. [#permalink] New post   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.
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Re: Given two intersecting circles.. [#permalink] New post   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
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Re: Given two intersecting circles.. [#permalink] New post   03 Dec 2018, 09:11
Sandy, how are we supposed to approximate square roots?
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Re: Given two intersecting circles.. [#permalink] New post   17 Dec 2018, 08:15
sandy wrote:
In this case let the circles intersect at B and F respectively, as shown in figure below:

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.
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Re: Given two intersecting circles.. [#permalink] New post   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:

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.. [#permalink] New post   25 Feb 2021, 08:21
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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:
\(\sqrt{2} = 1.41, \sqrt{3} = 1.73, \sqrt{5} = 2.23, \sqrt{7} = 2.64, \sqrt{10} = 3.16\)
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Re: Given two intersecting circles.. [#permalink] New post   22 Jul 2023, 11:01
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