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Consider a line L:3x+4y10=0L:3x+4y-10=0.

Consider a circle S1:x2+y216x18y+109=0S_1:x^2+y^2-16x-18y+109=0.

Find the equation of a certain circle S2S_2 which satisfies the 3 conditions below:

  • S2S_2 touches S1S_1 externally.

  • S2S_2 touches LL.

  • Line joining the centers of S1S_1 and S2S_2 is perpendicular to LL.

Do post nice solutions and hints!

Note by Nihar Mahajan
4 years, 4 months ago

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@Otto Bretscher @Tanishq Varshney @Chew-Seong Cheong @Jon Haussmann Please post hints/solutions.

Nihar Mahajan - 4 years, 4 months ago

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The method i suggest is that, the common tangent of the circles touching externally is the radical axis.

let S2=x2+y2+2gx+2fy+c=0S_{2}=x^2+y^2+2gx+2fy+c=0

Given S1=x2+y216x18y+109=0S_{1}=x^2+y^2-16x-18y+109=0

Radical axis of these circle is LL

S1S2=LS_{1}-S_{2}=L

compare the coefficient of xx and yy and constant.

Hope it helps

Tanishq Varshney - 4 years, 4 months ago

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actually the conditions u have put up are satisfied by the radical axis of circles

Tanishq Varshney - 4 years, 4 months ago

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Actually if you calculate out then it will come out that L does not touch S1 so it can not be the radical axis for the two touching circles.

Satvik Choudhary - 4 years, 4 months ago

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@Satvik Choudhary I had the same doubt. Tanishq bhayya wants to say that if we "construct" the radical axis and compare it with LL we get the requirements for equation of required circle.

Nihar Mahajan - 4 years, 4 months ago

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@Nihar Mahajan I tried to post my solution. I am having problem with the image uploading.

Satvik Choudhary - 4 years, 4 months ago

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@Satvik Choudhary What problem are you having with uploading the pic?

Nihar Mahajan - 4 years, 4 months ago

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@Satvik Choudhary Can u make a figure to justify ur statement, well i will post mine in few minutes

Tanishq Varshney - 4 years, 4 months ago

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Sorry to say but it is not necessary that S1S_1 touches LL. :(

Nihar Mahajan - 4 years, 4 months ago

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but radical axis is a line perpendicular to the line joining their centres

Tanishq Varshney - 4 years, 4 months ago

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@Tanishq Varshney Oh! I get it. So the strategy is to construct a radical axis , and since the slopes are equal , we can get the radius of required circle by comparing the equation of the lines.

Nihar Mahajan - 4 years, 4 months ago

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I think satvik's way is easier. While solving by your method , I encountered difficulties in computing cc which gave rise to a huge quadratic and I was f**ked! Anyway , I appreciate this idea of radical axis.

Nihar Mahajan - 4 years, 4 months ago

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The equation of S1 can be rewritten as (x8)2+(y9)2=(6)2(x-8)^{2}+(y-9)^{2}=(6)^{2}

The distance of the line L from center of S1 is

3(8)+4(9)1032+42=10\frac {3 (8)+4 (9)-10}{\sqrt {3^{2}+4^2}}=10

And since the radius of S1 is 6 The minimum distance between the S1 and the line is 4 which can also be seen as the diameter of the touching circle S2. Therefore the radius of S2 is 2. Now we have to find its centre.

The slope of the line perpendicular to L which is also the line joining their centers is 4/3.

Now using the parametric form

x83/5=y94/5=8\frac {x-8}{3/5}=\frac {y-9}{4/5}=-8

Since the distance between the two centers is (6+2)=8 and - sign can be clear from the diagram.

x=16/5 y=13/5 the coordinates of the center of S2.

Equation is

(x16/5)2+(y13/5)2=(2)2.(x-16/5)^{2}+(y-13/5)^{2}=(2)^{2}.

I might have done several mistakes please notify me if you find any. Hope I am right?

Satvik Choudhary - 4 years, 4 months ago

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Absolutely correct!

Alternate way : Find intersection of LL and the line joining centers. And use section formula with ratio 4:14:1 to obtain center.I obtained radius in the same way as yours.

Thanks @satvik choudhary !

Nihar Mahajan - 4 years, 4 months ago

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How to upload images i was just able to provide a link to it.

Satvik Choudhary - 4 years, 4 months ago

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@Satvik Choudhary !<no space>[<title>]<no space>(<link>)

Don't type: no space , title , link

Nihar Mahajan - 4 years, 4 months ago

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! The diagram

Satvik Choudhary - 4 years, 4 months ago

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