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Analytical Geometry

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Joined:	9 Sep 2007
Post:	130

10 Nov 2007 09:52:27 IST

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another circle ques plzzzz help

None

the locus of the mid point of a chord of the circle x^2 +y^2=4 which subtends a right angle at the origin

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Nikhil Gupta

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Joined: 23 Oct 2007

Posts: 279

10 Nov 2007 10:21:01 IST

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Let the midpoint of the chord be (h,k)

centre is(0,0)

distance of chord from centre is

2^2+2^2 = 8 (apply pythagoras theorem as chord subtends right angle and radius is 2)

by distance formula

(h-0)^2+(k-0)^2 =

squaring

h^2+k^2=8

replacing by x and y

locus is x^2+y^2=8

please rate if correct

venkat v.t

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Posts: 244

10 Nov 2007 13:15:13 IST

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write the formula that the equn., of chord having midpoint P(x1,y1)

is S1= S11

x x1+ y y1= x1^2+y1^2........(1)

Then homoginise the given circle with the help of (1)

Then use coeff of x^2+ coefft., of y^2 =0

we get x^2 + y^2 =1/2 is the locus.

'',' Off to a nu start!!

Blazing goIITian

Joined: 13 Jan 2007

Posts: 1289

10 Nov 2007 13:32:16 IST

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WELL I HV ANOTHER SOLN..

figure is shown below:-

from pythagoras only this qn. can be solved..

let AB be the chord and D (x,y) be mid-point..here c is origin...

from PT in

ABC

we get

AB^2 = 4+4

AB = 2

nopw as D is mid-point of AB

so we get

AD=BD =

now in

ACD (OR

BCD)

applying PT (pythagoras theorem) we get

AC^2 = AD^2 + CD^2

CD =

now appying distance formulae we get

CD =

(x-0)^2 + (y-0)^2

2 =

x^2 + y^2

hence req. locus is

x^2 + y^2 = 2

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