Question

If the angle between the tangent from $(0,0)$ to the circle $x^2+y^2+10x+10y+40=0$ is ${\tan }^{-1}\left(m\right)$, then $m=$

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{4}{3}$

Answer 1

The correct option is D $\frac{4}{3}$

$\begin{array}{c}{x}^2+y^2+10x+10y+40=0\\ {\left(x+5\right)}^2+{\left(y+5\right)}^2=0\\ R^2=10\\ now,\\ \theta ={\tan }^{-1}\left(\frac{2RL}{L^2-R^2}\right)\\ ={\tan }^{-1}\left(\frac{2\sqrt{10}\sqrt{40}}{30}\right)\\ \Rightarrow L^{\prime }=\sqrt{40}\Rightarrow L^2=40\\ ={\tan }^{-1}\left(\frac{4\times 10}{30}\right)\\ ={\tan }^{-1}\left(\frac{4}{3}\right)\\ m=\frac{4}{3}\\ hence,theoptionDisthecorrectanswer.\end{array}$