Paragraph: The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points…

Paragraph: The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$. Question: Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
  1. $2 x-\sqrt{5} y-20=0$
  2. $2 x-\sqrt{5} y+4=0$
  3. $3 x-4 y+8=0$
  4. $4 x-3 y+4=0$

Solution

Equation of tangent to hyperbola having slope $m$ is $ y=m x+\sqrt{9 m^2-4} $ Equation of tangent to circle is $ y=m(x-4)+\sqrt{16 m^2+16} $ Eqs. (i) and (ii) will be identical for $m=\frac{2}{\sqrt{5}}$ satisfy. $\therefore$ Equation of common tangent is $2 x-\sqrt{5} y+4=0$

Asked in: JEE Advanced 2010 (Paper 1)

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