IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

yachit

Two circles are drawn through the points (a, 5a) and (4a, a) to touch the axis of y. Prove that they intersect at an angle ${tan}^{-1}(\frac{40}{9})$ . [From the book: "Elements of co-ordinate Geometry" by S.L. Loney.]

shikhar_destructo

shikhar_destructo

First of all let us write the eq of family of circles through (a,5a) & (4a,a)

(x-a)(x-4a)+(y-5a)(y-a) +k(4x+3y-19a)=0 ( ie circle passing thru the intersection of circle with these 2 pts as the diameter and the line passing thru the 2 pts - What I mean is that any circle passing through the 2 pts. )

which simplifies to x^2+(4k-5a)x+y^2+93k-6a)y+9a^2-19ak=0

Now for any circle touching the y axis the radius^2 = (x cordinate of the centre of circle)^2

which gives 9k^2+76ak+36a^2=0 The 2 values of k represents the 2 circles passing thru the 2 pts and touching the y axis.
k1*k2=4a^2 and k1+k2= (5a-4k)

differentiating x^2+(4k-5a)x+y^2+93k-6a)y+9a^2-19ak=0 we get (2x+4k-5a)/(6a-2y-3k)=(dy/dx)

substituting (a,5a) we get (dy/dx)= (3a-4k)/(4a+3k) for 2 values k1 & k2

so angle of intersection is tan-1(((3a-4k1)/(4a+3k1)-(3a-4k2)/(4a+3k2))/(1+(3a-4k1)(3a-4k2)/((4a+3k1)(4a+3k2)))

replace k1k2 & k1+k2 with their values and get tan-t(4/9) I got it so ans comes verified.

I think I do deserve a rate for this :)

shikhar_destructo

Sorry i wrote tan-1(4/9) in place of tan-1(40/9) carelessly while typing.

krishna.gopal

Sorry i have not solved it myself but approach of shikhar is perfect. And Shikhar you definetly deserve more rates for this reply.

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