Find range of $a$ such that angle between pair of tangents from $(a,a)$ to a circle lies within $(\frac{\pi}{3},\pi)$

Question

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

Answer 1

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 $a\in (-3,-1)\cup (3,5)$.

Answer 2

Observations: the point $(a,a)$ we seek is on the line $y=x$, which has slope 1, or angle $\pi/4$. The equation of the circle is $(x-1)^2+(y-1)^2=8$, whose center lies on this line. We'll solve the problem on the 'right side' of the circle and use symmetry to figure out the 'left side'. By symmetry, the condition, "the angle between the pair of tangent lines is in $(\pi/3,\pi)$," is equivalent to the condition, "the angle between the tangents and $y=x$ is in $(\pi/6,\pi/2)$".

In other words, we want the lines that intersect $y=x$ and have slope $\tan(\pi/4\pm \pi/6)$. We have $\tan(5\pi/12)=2+\sqrt{3}$, $\tan(\pi/12)=2-\sqrt{3}$. We can find where on the circle the tangent lines have these slopes: they are at $(2+\sqrt{3},2-\sqrt{3})$ and $(2-\sqrt{3},2+\sqrt{3})$, respectively. Now we can find the first point with the required angle: it'll be where one of these lines intersects $y=x$. $$ (2+\sqrt{3})(x-(2+\sqrt{3}))+(2-\sqrt{3})= y = x;\qquad (x,y)=(5,5) $$So the viable options are $3<a<5$. By symmetry, we have $-3<a<-1$ as well.