Question

If $\theta$ is the angle between the tangents from $(-1,0)$ to the circle $x^2+y^2-5x+4y-2=0$, then $\theta$ is equal to:

• A. $2{\tan }^{-1}\frac{7}{4}$
• B. ${\tan }^{-1}\frac{7}{4}$
• C. $2{\cot }^{-1}\frac{7}{4}$
• D. ${\cot }^{-1}\frac{7}{4}$

Answer 1

The correct option is A

$2{\tan }^{-1}\frac{7}{4}$

Explanation for the correct option.

The general equation of a circle is $x^2+y^2+2gx+2gy+c=0$, and its Centre is $\left(-g,-f\right)$ and radius is $\sqrt{g^2+f^2-c}$.

The angle between two tangents from a point $\left(x_1,y_1\right)$ and a circle $x^2+y^2=r^2$ is $2{\tan }^{-1}\frac{r}{\sqrt{S_1}}$.

The equation of circle is given to be $x^2+y^2-5x+4y-2=0$.

By comparing it with the general equation of circle, we get $g=-\frac{5}{2},f=2,c=-2$.

So, the Centre $O$ of the circle will be, $\left(\frac{5}{2},-2\right)$ and the radius will be $\frac{7}{2}$.

The coordinates of the given point are $(-1,0)$.

Substituting the value in given equation of circle, we get

$\begin{array}{rcl}{S}_1& =& {\left(-1\right)}^2+{\left(0\right)}^2-5\left(-1\right)+4\left(0\right)-2\\ & =& 4\end{array}$

So,

$\begin{array}{rcl}\theta & =& 2{\tan }^{-1}\left(\frac{{\displaystyle \frac{7}{2}}}{\sqrt{4}}\right)\\ & =& 2{\tan }^{-1}\left(\frac{7}{4}\right)\end{array}$

Hence, option (A) is correct.