JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Circle

The two circles are $x^2+y^2=4$ (center $(0,0)$, radius 2) and $(x-3{)}^2+y^2=1$ (center $(3,0)$, radius 1).

The distance between centers is 3, and $3=2+1$, so the circles are externally tangent. The point of tangency is at $(2,0)$.

Direct common tangents meet at the external center of similitude, which divides the line joining centers externally in the ratio of radii $2:1$. This point is at $(6,0)$.

A tangent through $(6,0)$ to $x^2+y^2=4$ has the form $y=m(x-6)$. The distance from origin to this line equals 2:






For $m=-\frac{1}{2\sqrt{2}}$: $y=-\frac{1}{2\sqrt{2}}(x-6)$, or $2\sqrt{2}y=-x+6$, giving $x+2\sqrt{2}y=6$.