IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

rahul_c

C1 and C2 are 2 concentric circles ,the radius of C2 being twice that of C1.From a point P on C2,tangents PA and PB are drawn to C1 touching
at A and B.Prove that the centroid of the triangle PAB lies on the circle C1

malay

this is a hint
Let PA touches C1 at M
then M is the mid-point of PA

rahul_c

malay if you can kindly elaborate

malay

can you just give me the equation of the tangents from the point(x,y) to the circle of radius r with centre at (0,0)

magiclko

the centroid of the triangle will be the mid-point of the line joining the common centre and P, and hence will lie on C1

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Manasi....

Life's tuff, but m tuffer.........

malay

magiclko
if PF is median, and Gis centroid, then:
PG:GF=2:1

magiclko

yupp....u r right malay....

PF is the median for the triangle PAB, n will defintely be divided by the centroid G in ratio 2:1.... but the point G will be the centre of the circle passing thru points A, B, P and O (the common centre of the concentric circles). as u said if PF is median, then PF produced will pass thru O, and GF= OF and also PG = GO = R and thus PG:GF = 2:1....

Hence Proved.....

malay

here is the solution to it:

Join AB and OP(O is the common centre of two circles). Let they intersect at M.

Let PO intersect the inner circle at E.

Now, PAB is an isoceles triangle(PA=PB). Hence perpendicular from P will also bisect AB.

Perpendicular from O to AB will also bisect AB.

From the above two statements, OP is perpendicular bisector of AB, hence M is the mid-point of AB.

Now, consider right triangle AOP.
OA=r
AP^2=OP.(OP+2r)=r*3r{tangent -secant theorem}
AP=r*square root of 3
hence PO=2r

Now, AM is perpendicular to OP
hence, AM.OP=AP.AO
AM^2=3r^2/4

In right triangle OAM
OM^2=OA^2-AM^2=r^2/4
OM=r/2

hence EM=r/r

Now, PM is median and PE/EM=2/1
hence E is centroid.

rahul_c

good job malay

krishna.gopal

There is another way to look at the solution. Make the diagram and use points used by Malay. (ie O is common center, PA and PB are tangents. OP cut inner circle at E and AB at B)

PAO is a right angle triangle with OP=2r and OA = r. Therefore angle (APO)=sin^-1(1/2)=30^o. and hence angle (APB)=60^o. PA=PB means its isoceles as well. Thus its a equiletral triangle of silde

3r. PM is perpendicular median with length 3r/2. Thus PE=r divides median in ratio of 2:1 and hence centroid

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