I have to find the exhaustive range of values of $a$ such that the angle between the pair of tangents drawn from $(a,a)$ to the circle
$$x^2+y^2-2x-2y-6=0$$
lies in the range $(\frac{\pi}{3},\pi)$.
So I tried this as follows:
I know that the angle between tangents from a point $(x_1,y_1)$ is $\theta = 2\arctan(\frac{r}{\sqrt S_1})$, where $r$ = radius of the circle, and $\sqrt S_1 =$ $x_1^2+y_1^2+2gx_1+2fy_1+c$ if the equation of the circle is $x^2+y^2+2gx+2fy+c=0$.
Using this I tried but cant get the upper and lower bounds for the value of $a$. Help would be appreciated
Best Answer
The standard circle form is $(x-1)^2+(y-1)^2=r^2=(2\sqrt2)^2$. Let $O(1,1)$, $A(a,a)$ and the subtended angle $\theta$. Then, $|AO|=\sqrt{2(a-1)^2}=\pm\sqrt2(a-1)$ and
$$ \sin \frac{\theta}2 = \frac r{|AO|} = \frac{2\sqrt2}{\pm \sqrt2(a-1)} =\pm \frac2{a-1}$$
Given that
$$ \frac12 =\sin\frac\pi6< \sin \frac{\theta}2 < \sin \frac\pi2 =1$$
we have
$$ \frac12 < \pm \frac2{a-1}<1$$
which leads to the ranges $a\in (-3,-1)\cup (3,5)$.