Question

The angle between the pair of tangents from the point $ \left( 1,\,\,\frac{1}{2} \right) $ to the circle $ {{x}^{2}}+{{y}^{2}}+4x+2y-4=0 $ , is

• A. $ {{\cos }^{-1}}\left( \frac{4}{5} \right) $
• B. $ {{\sin }^{-1}}\left( \frac{4}{5} \right) $
• C. $ {{\tan }^{-1}}\left( \frac{4}{5} \right) $
• D. None of these

Answer 1

Answer: (B)

The equation of pair of tangent from the point
$ \left( 1,\,\,\frac{1}{2} \right) $ to the circle
$ S\equiv {{x}^{2}}+{{y}^{2}}+4x+2y-4=0 $ is $ S{{S}_{1}}={{T}^{2}} $
$ ({{x}^{2}}+{{y}^{2}}+4x+2y-4)\left( 1+\frac{1}{4}+4+1-4 \right) $
$ ={{\left[ x+\frac{1}{2}y+2(x+1)+y+\frac{1}{2}-4 \right]}^{2}} $
$ \frac{9}{4}({{x}^{2}}+{{y}^{2}}+4x+2y-4)={{\left( 3x+\frac{3}{2}y-\frac{3}{2} \right)}^{2}} $
$ \Rightarrow $ $ \frac{9}{4}({{x}^{2}}+{{y}^{2}}+4x+2y-4)=\frac{9}{4}{{(2x+y-1)}^{2}} $
$ \Rightarrow $ $ {{x}^{2}}+{{y}^{2}}+4x+2y-4 $
$ =4{{x}^{2}}+{{y}^{2}}+1+4xy-2y-4x $
$ \Rightarrow $ $ 3{{x}^{2}}+4xy-8x-4y+5=0 $
On comparing this equation with
$ a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 $ , we get
$ a=3,\,\,h=2,\,\,b=0,\,\,g=-4,\,\,f=-2,\,\,c=5 $
Required angle $ ={{\tan }^{-1}}\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right| $
$ ={{\tan }^{-1}}\left| \frac{2\sqrt{4-0}}{3} \right| $
$ ={{\tan }^{-1}}\frac{4}{3} $
$ ={{\sin }^{-1}}\left( \frac{4}{5} \right) $