Determine the locus of $A$ when $B$ traverses the circle $\mathcal{C}$.

Question

Can someone help me to solve this exercice:

Consider a circle $\mathcal{C}$ with centre $O$ and radius $2$ and a point $P$ such that distance $OP=1$. Let $B$ be a point on $\mathcal{C}$ and let $A$ be the point on the line $(OB)$ equidistant from $P$ and $B$. Determine the locus of $A$ when $B$ traverses the circle $\mathcal{C}$.

My attempt:

First I define the points $P(1;0)$ and $B(2\cos{\theta}, 2\sin{\theta})$ without loss of generality. The point that lies in the middle of $P$ and $B$ is $I(\frac{1+2\cos{\theta}}{2}, \sin{\theta})$.

Next I tried to find the equation of the line $(BP)$:

$(BP): x + \frac{1-2\cos{\theta}}{2\sin{\theta}}y=1$

Then the vector

$\vec{n} = (1, \frac{1-2\cos{\theta}}{2\sin{\theta}})$ is normal to the line $(BP)$.

An equation of the line that goes trough the point $I$ and is normal to $(BP)$ is:

$ d \equiv y = \frac{1-2\cos{\theta}}{2\sin{\theta}}x+\frac{3}{4\sin{\theta}}$

Finally the only thing to do is to find the intersection of $d$ and $OB$ to find the coordinates of $A$:

$(OB) \cap d: \left(\frac{3}{2} \frac{\cos{\theta}}{2-\cos{\theta}}, \frac{3}{2} \frac{\sin{\theta}}{2-\cos{\theta}}\right)$

My problem now is to see what curve it is if $B$ (thus $\theta$) varies. (I know because of GeoGebra that it is an ellipse, but how can you see this only from the coordinates of the intersection point?)

Answer 1

Diving by $\cos \theta$

$$ A(x,y)\equiv\left(\frac{3}{4\sec\theta-2},\frac{3\tan\theta}{4\sec\theta-2}\right) \\ \ \\ \sec\theta=\left(\frac{3+2x}{4x}\right)\\ \ \\ \begin{align} y^2&=\frac{\left(9\tan^{2}\theta\right)}{\left(16\sec^{2}\theta+4-16\sec \theta\right)}=\frac{\left(9\sec^{2}\theta-9\right)}{\left(16\sec^{2}\theta+4-16\sec \theta\right)}\\&=\frac{9\left(\left(\frac{3+2x}{4x}\right)^{2}-1\right)}{\left(16\left(\frac{3+2x}{4x}\right)^{2}+4-16\left(\frac{3+2x}{4x}\right)\right)}=-\frac{3}{16}\left(4x^{2}-4x-3\right) \end{align} $$

The final locus is $$ \left(x-\frac{1}{2}\right)^{2}+\frac{4y^{2}}{3}=1 $$

The locus happens to be an ellipse with focii at O,P we can use this to find locus using pure geometry instead of analytical geometry

Let $\mathcal{C}_1$ be a circle centered at $O$ with radius R and $\mathcal{P}$ be a point on circle $\mathcal{C}_2$ with radius r (R>r)

$$ R=OB\\ R=OA+AB=OA+PA $$

This shows that the locus is an ellipse with focii $O,\mathcal{P}$ with eccentricity $e=\frac{r}{R}$

Answer 2

HINT: $$AO+AP=AO+AB=OB=2.$$ Hence...