Circles

A circle with radius a and centre on y - axis slides along it and a variable line through (a,0) cuts the circle at points P and Q. What is the region in which the pt of intersection of tangents to the circle at points P and Q lies ?

16 Answers

11
pawanseerwani seerwani ·

hey guys,
let the circle be x2+(y-k)2=a2
let (p,q) be a point on the required locus
from data xp+(y-k)(y-q)=a2(chord of contact from (p,q))
this passes through (a,0)
thus ap+k(k-q)=a2
ie k2-qk=ap-a2=0
k is real hence discriminant>0
hence u get q2-4a(p-a)>0
hence the region is given by

Y2-4a(X-a)>0
it lies outide the obtained parabola

NOTE: LET ME KNOW IF I AM CORRRECT OR NOT(PLEASE)

1
Optimus Prime ·

this will the diagram

62
Lokesh Verma ·

yes this will be the diagram.. and the proof?

1
Optimus Prime ·

ya wait , i am thinking

1
Optimus Prime ·

what does it mean by what is region in which the point of intersection lies??

1
skygirl ·

i thnk it means like the line through a,0 can be a tangent on either side... they are the limitinng cases...

oh//i cant express... trying to draw...

1
skygirl ·

62
Lokesh Verma ·

I think we cando better than that sky....

because the location of the circle on the z axis is unknown..

1
skygirl ·

z?
ok u mean y //?

1
Philip Calvert ·

wat about" on y - axis slides along it "

1
skygirl ·

phir toh poora ho jaega... poora plane...

1
Philip Calvert ·

wahi toh aisa nahi hona chahiye

1
Philip Calvert ·

jaise jaise circle dur jaega dikkat hogi..[12]
jab origiin pe aaega or fir neeche jaega [12]

1
skygirl ·

[12]

arey poora plane hi cover ho jaara....

1
mithun ·

base AB(=k units) of a triangle ABC is fixed ,vertex C moves in such a way that
sinA=θsinB (|θ|<1) ,
then locus of C is a circle whose center lies on base AB ,find the radius of this circle?

1
Vivek ·

yes u're right

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