IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - Please solve some JEE questions from circles for me...

sahilgupta_iit

Q.1. Through a fixed point (h,k) secants are drawn to the circle x^2 + y^2 = r^2. Show that the locus of the mid-points of the secants intercepted by the circle is x^2 + y^2 = hx + ky. [IIT-JEE 1983]

Q.2. Let a given line L1, intersect x and y axos at P and Q respectively. Let another line L2 , perpendicular to L1, cut the y axis at R and S respectively. Show that the locus of the point of intersection of the lines PS and QR is a circle passing through the origin. [1987]

Q.3. Let S --- x^2 +y^2 +2gx + 2fy + c = 0 be a given circle. Find the locus of the foot of perpendicular drawn from the origin upon any chord of S which subtends a right angle at the origin. [1988]

Q.4. If (mi, 1/mi), mi > 0, i = 1,2,3,4 are the four distinct points on a circle, then show that m1*m2*m3*m4 = 1. [IIT-JEE MAINS, 1989]

Q.5. Let a circle be given by 2x(x-a) + y(2y-b) = 0 (a,b not equal to 0), find the condition on a and b if two chords, each bisected by the x axis, can ce drawn to the circle from (a, b/2). [1992]

Rates assured.

Conjurer

1) The locus of the chord of a circle whose midpoint is given is S = T

Use this in the first case taking x1,y1 as the midpoint:

x1^2 + y1^2 - r^2 = xx1 + yy1 -r^2

This gives x1^2 +y1^2 = xx1+yy1

This line passes through (h,k)

SO putting x= h and y=k

We get the locus as x1^2 +y1^2 = hx1 +ky1

Writing in general form we get x^2 + y^2 = hx + ky

sahilgupta_iit

Common guys solve the other questions also.

anchitsaini

2)eqn of L1=x cos

+ y sin

=p1

eqn of L2=x sin

- y cos

=p2

from above 2 eqns--

P--p1sec

, 0

Q--0,p1cos

R--p2 cos

,0

S--0,-p2sec

eqn of PS--

comes to be

x/p1 - y/p2=sec

--------1

eqn of QR--

comes to be

x/p2 + y/p1=cosec

---------2

locus of pt of intersection is obtained by eliminating p2

from 1 and 2

it comes to be--

x^2 + y^2 - p1x sec

- p1ycosec

=0

as c=0 in the above eqn

hence the circle passes through origin

PROVED

anchitsaini

4)
let the circle be x^2 + y^2 + 2gx + 2fy + c=0

putting

x=mi

y=1/mi

in the above eqn we get

mi^2 + 1/mi^2 + 2gmi + 2f/mi + 1=0

or

mi^4 + 2gmi^3 + cmi^2 + 2fmi +1=0

this eqn is of degree 4

hence it has 4 roots-m1,m2,m3,m4

product of roots

=m1m2m3m4=1/1=1

PROVED

sahilgupta_iit

Carry on guys. Solve more.

anchitsaini

these are iit qns man
don't expect the answers so soon!!!

anchitsaini

5)
eqn of circle comes to be

x^2 + y^2 - ax -by/2=0

let the chord be bisected at (h,0) since it is bisected by

the x axis

the eqn of chord would be found as

we know the middle pt

hence eqn is T=S1

from this the eqn of chord comes to be--

(h-a/2)x - by/4 - ah/2 + ah - h^2=0

since this chord passes through (a,b/2) hence this should

satisfy the eqn of chord--

ie

(h-a/2)a - b*b/2*4 - ah/2 + ah - h^2=0

the final eqn is

h^2 - 3ah/2 + a^2/2 + b^2/8 =0 ----------1

FROM HERE ON I AM NOT SURE--

since two chords can be drawn therefore

the eqn 1 should have D>0

9a^2/4 - 4(a^2/2 + b^2/8)>0

hence

a^2/4 - b^2/2>0

ans seems to be

a^2>2b^2

i hope it is correct

anchitsaini

enough for now!!!!!!!
i need rates for this, seriously!!!
:)

sahilgupta_iit

kudos to you all.

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