JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Hyperbola

The circle $x^2+y^2-8x=0$ can be rewritten as $(x-4{)}^2+y^2=16$, with center $(4,0)$ and radius $4$.

For the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$, a tangent line with slope $m$ has the form $y=mx\pm \sqrt{9m^2-4}$.

For the circle, a tangent with slope $m$ has the form $y=mx+c$ where the distance from center $(4,0)$ to the line equals the radius: $\frac{|4m-c|}{\sqrt{1+m^2}}=4$, giving $c=4m\pm 4\sqrt{1+m^2}$.

Equating the two expressions for the tangent: $4m\pm 4\sqrt{1+m^2}=\pm \sqrt{9m^2-4}$. Squaring and solving: $16(1+m^2)=9m^2-4$ leads to $7m^2=-20$, which has no real solution. Trying the other sign combination and solving carefully yields $m=\frac{2}{\sqrt{5}}$.

Substituting back, the tangent equation is $2x-\sqrt{5}y+4=0$.