IIT-JEE 2010 · Paper 1, Mathematics question 45

Tangents to the hyperbola $\frac{x^2}9-\frac{y^2}4=1$: $y=mx\pm\sqrt{9m^2-4}$.

Check (B): $y=\frac2{\sqrt5}x+\frac4{\sqrt5}$. Here $9m^2-4=\frac{36}5-4=\frac{16}5$, so $c=\frac4{\sqrt5}$ fits. For the circle (centre $(4,0)$, radius 4): distance $\frac{|8+4|}{\sqrt{4+5}}=4$. It touches both, and its slope is positive.

Others. (A) is at distance $\frac{12}3=4$ from $(4,0)$ but $c^2=80\ne\frac{16}5$; (C) and (D) are not tangent to the hyperbola ($9m^2-4$ does not match).

(B)