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 $S{S}_1=T^2$
$(x^2+y^2+4x+2y-4)\left(1+\frac{1}{4}+4+1-4\right)$
$={\left[x+\frac{1}{2}y+2(x+1)+y+\frac{1}{2}-4\right]}^2$
$\frac{9}{4}(x^2+y^2+4x+2y-4)={\left(3x+\frac{3}{2}y-\frac{3}{2}\right)}^2$
$\Rightarrow$ $\frac{9}{4}(x^2+y^2+4x+2y-4)=\frac{9}{4}{(2x+y-1)}^2$
$\Rightarrow$ $x^2+y^2+4x+2y-4$
$=4x^2+y^2+1+4xy-2y-4x$
$\Rightarrow$ $3x^2+4xy-8x-4y+5=0$
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 $={\tan }^{-1}\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$
$={\tan }^{-1}\left|\frac{2\sqrt{4-0}}{3}\right|$
$={\tan }^{-1}\frac{4}{3}$
$={\sin }^{-1}\left(\frac{4}{5}\right)$