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Analytical Geometry

Blazing goIITian

 Joined: 10 Oct 2011 Post: 430
21 Jul 2012 12:48:58 IST
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Let S1=x2+y2-4x-8y+4=0 and S2 be its image in the line y=x, find the equation of the circl
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Analytical Geometry

Let S1=x2+y2-4x-8y+4=0 and S2 be its image in the line y=x, find the equation of the circle touching y=x at (1,1) and its radical axes with S2 passes through the centre of S1

#### Comments (1)

New kid on the Block

Joined: 24 Jul 2012
Posts: 14
28 Jul 2012 13:51:05 IST
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w.r.t line y=x image of S1 i.e., S2=x2+y2-8x-4y+4=0

let req circle be S3=x2+y2+2gx+2fy+c=0

pt(1,1) lies on S3

2g+2f+c+2=0-----------(1)

centre of S3 is (-g,-f)

slope of y=x is 1

so slope of line joining (1,1)&(-g,-f) is -1 (perpendicular)

so g+f+2=0------------(2)(line joining centre &(1,1))

radicle axis will be S2-S3=0

centre of S1 i.e., (4,2)lies on radicle axis

so 4g-8f-c-28=0---------------(3)

solve (1),(2)&(3) & get values of g,f,c& substitute in S3 u will get the req circle.

by the way thanks 4 a good problem

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