Locus of a point from which tangents drawn to a fixed circle are inclined at a constant angle $\theta$ and area bounded by it

Question

The question is given above. I'm able to get the locus, however, I'm unable to comment on the area bounded by it.

We can get the locus using the relation $\tan\theta=\frac{2LR}{L^{2}-R^{2}}$ where $R$ is radius of circle and $L$ is length of tangent from the given point. We can write $L=\sqrt{h^{2}+k^{2}-25}$. On solving we get the locus as $$\tan^{2}\theta(x^{2}+y^{2}-50)^{2}=100(x^{2}+y^{2}-25)$$ How do we comment on the area bounded by this curve for $\theta\in (\frac{\pi}{2},\pi)$ and $\theta\in(\frac{\pi}{4},\frac{\pi}{2})$. Alternately, is there another way to see the area bounded by the locus of such a point without finding the locus.

Answer 1

The equation of the locus you have found looks exact, but it's hard for me to find out its origin (I don't understand your $\tan\theta=\frac{2LR}{L^{2}-R^{2}}$ and how I can exploit it).

But there is a simpler approach.

If $M$ is any point in the plane, let $T$ be one of the two points of tangency of a tangent issued from $M$ to the circle with center $O$ and radius $5$. In right triangle $MTO$, we have

$$MO=\frac{5}{\sin(\theta/2)}$$

Therefore, for a constant $\theta$, i.e., for a constant $\sin(\theta/2)$, we have a constant distance $MO$ : the locus of such points $M$ is the circle with center $0$ and radius

$$r=\frac{5}{\sin(\theta/2)}$$

with area

$$A(\theta)=\pi \frac{25}{\sin^2(\theta/2)}$$

As $A(...)$ is a decreasing function of $\theta$ on $[(\pi/4),\pi]$, you just have to compute the value of $A$ at the lower bound on each interval, for example, in the third case, $[(\pi/4),(\pi/2)]$, just compute $A(\pi/4)$.

Can you take it from here ?

Answer 2

Can you make a helpful sketch to find what is the area referred to in your question, between the circle and tangents,two polar angles and the curve?

I am able to recognize the two tangents and triangles with sides adjoining right angle at tangent point as $ R=5,L,h,k $ and hypotenuse $$ \sqrt{25 +L^2} = \sqrt{h^2+k^2}. $$

And since $ \tan \frac\theta2= \frac 5L=\frac RL=t, $ you calculated

$$ \tan \theta= \frac{2t}{1-t^2} = \frac{2RL}{L^2-R^2}$$