# Help required.

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

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

Find the equation of a certain circle $S_2$ which satisfies the 3 conditions below:

• $S_2$ touches $S_1$ externally.

• $S_2$ touches $L$.

• Line joining the centers of $S_1$ and $S_2$ is perpendicular to $L$.

Do post nice solutions and hints! Note by Nihar Mahajan
5 years, 4 months ago

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

The distance of the line L from center of S1 is

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

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

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

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

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

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

- 5 years, 4 months ago

- 5 years, 4 months ago

Absolutely correct!

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

Thanks @satvik choudhary !

- 5 years, 4 months ago

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

- 5 years, 4 months ago

Don't type: no space , title , link

- 5 years, 4 months ago

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

let $S_{2}=x^2+y^2+2gx+2fy+c=0$

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

Radical axis of these circle is $L$

$S_{1}-S_{2}=L$

compare the coefficient of $x$ and $y$ and constant.

Hope it helps

- 5 years, 4 months ago

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

- 5 years, 4 months ago

Sorry to say but it is not necessary that $S_1$ touches $L$. :(

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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.

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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.

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 months ago

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

- 5 years, 4 months ago

- 5 years, 4 months ago

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

- 5 years, 4 months ago