JEE Mains Model Paper 2

y = mx - b√1+m2 is a tangent to the circle x2+y2=b2 for all values of m. if it also touches thecircle (x−a)2+y2=b2 , then the length of the perpendicular from its centre (a,0) on this line is equalto the radius b of the circle, which gives.

ma−b√1+m2

√1+m2

= ±b
Taking negative value on R.H.S we get m=0, so we neglect it.
Taking the positive value on R.H.S we get
ma = 2b√1+m2
⇒ m2(a2−4b2) = 4b2
⇒ m =

2b

√a2−4b2