996writing 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 )
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)))
if we replace k1k2 & k1+k2 with their values and we get tan-(40/9)