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2 circles are tangent each other and be x-axis above. What is the slop

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2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   19 Mar 2016, 23:12
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2 circles are tangent each other and be x-axis above. What is the slope of the line that passes through 2 centers of 2 circles?

A. 2/√96
B. 1/√6
C. 1/2√3
D. 1/√3
E. 2√6


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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   19 Mar 2016, 23:51
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Slope = Change in Y/Change in X

Change in Y = 6 - 4 = 2

The length of line joining 2 centres = 6 + 4 = 10

Change in X = \(\sqrt{10^2 - 2^2}\) = \(\sqrt{96}\)

Slope = \(\frac{2}{\sqrt{96}}\)

Answer: A
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   20 Mar 2016, 06:47
Vyshak wrote:
Slope = Change in Y/Change in X

Change in Y = 6 - 4 = 2

The length of line joining 2 centres = 6 + 4 = 10

Change in X = \(\sqrt{10^2 - 2^2}\) = \(\sqrt{96}\)

Slope = \(\frac{2}{\sqrt{96}}\)

Answer: A



Can you please explain how you got Change in X = square root of 10^2-2^2
I thought change in X should be the distance between the two perpendiculars dropped from the centers of both the circles, and this difference would be 10. Please suggest what I am doing wrong.
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2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   20 Mar 2016, 07:01
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MeghaP wrote:
Vyshak wrote:
Slope = Change in Y/Change in X

Change in Y = 6 - 4 = 2

The length of line joining 2 centres = 6 + 4 = 10

Change in X = \(\sqrt{10^2 - 2^2}\) = \(\sqrt{96}\)

Slope = \(\frac{2}{\sqrt{96}}\)

Answer: A



Can you please explain how you got Change in X = square root of 10^2-2^2
I thought change in X should be the distance between the two perpendiculars dropped from the centers of both the circles, and this difference would be 10. Please suggest what I am doing wrong.


Consider a perpendicular line drawn from the centre of the smaller circle to the radius of the larger circle (Call it as 'a').

We know that hypotenuse = Distance between the line joining the two centres = 10

The other side = 2

By pythagoras theorem, \(a^2 + 2^2\) = \(10^2\) --> a =\(\sqrt{96}\)
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   20 Mar 2016, 07:07
Vyshak wrote:
MeghaP wrote:
Vyshak wrote:
Slope = Change in Y/Change in X

Change in Y = 6 - 4 = 2

The length of line joining 2 centres = 6 + 4 = 10

Change in X = \(\sqrt{10^2 - 2^2}\) = \(\sqrt{96}\)

Slope = \(\frac{2}{\sqrt{96}}\)

Answer: A



Can you please explain how you got Change in X = square root of 10^2-2^2
I thought change in X should be the distance between the two perpendiculars dropped from the centers of both the circles, and this difference would be 10. Please suggest what I am doing wrong.


Consider a perpendicular line drawn from the centre of the smaller circle to the radius of the larger circle (Call it as 'a').

We know that hypotenuse = Distance between the line joining the two centres = 10

The other side = 2

By pythagoras theorem, \(a^2 + 2^2\) = \(10^2\) --> a =\(\sqrt{96}\)



Thank you so much for a prompt response. This is very helpful :)
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   21 Mar 2016, 18:20
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2 circles are tangent each other and be x-axis above. What is the slope of the line that passes through 2 centers of 2 circles?

A. 2/√96
B. 1/√6
C. 1/2√3
D. 1/√3
E. 2√6

Attachment:
2 circles.jpg
2 circles.jpg [ 10.93 KiB | Viewed 3688 times ]

Just like the picture above, d=√96 is derived from d^2=(10^2)-(2^2)=96. Then, the slope of the line that passes through 2 centres=length of YZ/length of XZ. Thus, the slope is 2/√96. Therefore, the answer is A.
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   18 Nov 2016, 22:02
why do we need to find length between centers? We need to find slope of line

Slope=rise/run

rise=6-4=2
run=4+6=10

So, slope is 2/10=1/5
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   18 Nov 2016, 23:05
Temurkhon wrote:
why do we need to find length between centers? We need to find slope of line

Slope=rise/run

rise=6-4=2
run=4+6=10

So, slope is 2/10=1/5


Change in x is not 10. The centers of the two circles are different. You have to apply pythagoras theorem to determine the change in x.
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   18 Nov 2016, 23:16
change in X coordinates is the length of horizontal line from one point to other. Here it is 10
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   18 Nov 2016, 23:35
Temurkhon wrote:
change in X coordinates is the length of horizontal line from one point to other. Here it is 10


The radius of the two circles (4 and 6) forms a line whose slope has to be determined. You are assuming the combined radius of the line itself as the change in x value.
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   19 Nov 2016, 00:03
Ah, I see, circles make me crazy. Thanks
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   25 May 2020, 10:03
Slope = Change in Y/Change in X= 2

The length of line joining 2 centres = 6 + 4 = 10

Calculate, Change in X and Slope

Answer: A

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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink] New post   29 Aug 2021, 04:42
Hi Bunuel, could you please help me for this question? Thank you.
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Re: 2 circles are tangent each other and be x-axis above. What is the slop [#permalink]
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