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Analytical Geometry

Hot goIITian

Joined:	12 Mar 2007
Post:	189

4 Jun 2007 14:39:25 IST

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475

circle 29

None

The equation of circle which touches the axes of co-ordinates and the line x/3 + y4 = 1 & whose centre lies in the 1st quadrant is, x²+ y²- 2cx - 2cy + c²= 0, where c is :

(A) 1

(B) 2

(D) 6

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Comments (3)

T RAHUL N THIRUGNANAM

Cool goIITian

Joined: 23 Mar 2007

Posts: 85

4 Jun 2007 14:52:09 IST

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center of the circle is(c,c)
line is 4x+3y-12=0
radius=c.
c=(4c+3c-12)/5 or c=(-4c-3c+12)/5
c=6 or c=1
hence ans = 6,1(a,d) ate me plz

Sanchay Gupta

Hot goIITian

Joined: 12 Mar 2007

Posts: 189

6 Jun 2007 14:36:51 IST

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ur ans is not totally correct

in answers

c=6 but c is not equal to 1

joy francis

Blazing goIITian

Joined: 19 Feb 2007

Posts: 1802

6 Jun 2007 14:53:50 IST

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Hi.

The ans is 6.

The circle touches the axes and its center lies in the first quadrant

.: center has coordinates (r,r)

as per the equation given r=c

also, 4x+3y=12 is a tangent hence its distance from (r,r) should be r only.

.: r = 4c+3c-12/root(16+9)=7c-12/5

c=6

@rahulvedanta

distance of a point x1,y1 from a line ax+by+c=0 is just

ax+by+c/root(a^2+b^2)..it is not plusminus[ax+by+c/root(a^2+b^2)]

logically thinking it is also correct how can a given line and a given point have

two different distances between them

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