A locus question involving circles

Question

I had come across this question:

A line $OB$ meets the circle $$x^2+y^2-ax=0$$ at $B$. From $B$, draw a line $BC$ perpendicular to $x$-axis and draw a line $CM$ perpendicular to $OB$. Then, what is the locus of $M$?

I tried thinking of various ways to proceed, however I couldn't form any idea in my mind.

It would be helpful if someone could hint at an approach to this question.

Is there a better way rather than solving for each and every point?

Answer 1

This is a solution without trigonometry. The $x$-coordinate of point $B$, which is the intersection of the line $y=kx$ and the circle $y^2=ax-x^2$, is obtained by setting $kx=\sqrt{ax-x^2}$. This gives

$$X_B=\frac{a}{k^2+1}\\ $$

Since this is also the $x$-coordinate of $C$, we get that the coordinates of $C$ are

$$C\left(\frac{a}{k^2+1},0 \right)$$

The line $CM$ must have an angular coefficient equal to $-1/k$. Since it passes through $C$, we have $0=-1/k \cdot a/(k^2+1) + z$. This leads to $z=a/[k(k^2+1)]$. Thus, the line $CM$ has equation

$$y=-\frac{1}{k} x + \frac{a}{k(k^2+1)}$$

Now if we call $x,y$ The coordinates of $M$, they satisfy

$$y=-\frac{1}{k} x+ \frac{a}{k(k^2+1)} $$

and because $k=y/x$, substituting we have

$$ y^2 =\sqrt{a} x^{3/2} - x^2$$

This is the graph of the locus for $a=5$.

Answer 2

It's rather clear that the given equation is equation of the circle, centered at $(\frac{a}{2},0)$ with radius $\frac{a}{2}$: $$x^2+y^2-ax=0\Leftrightarrow \left(x-\frac{a}{2}\right)^2+y^2=\frac{a^2}{4}.$$ It's known that a circle centered at $(a,b)$ with radius $R$ can be parametrized with $$\begin{cases} x=a+R\cos\varphi\\ y=b+R\sin\varphi \end{cases}.$$
So we scale the entire thing down by $\frac{a}{2}$, do the calculations and scale by $\frac{a}{2}$ back up.

So let's start the derivation.
Let $a=2$, let $B(1+\cos\varphi,\sin\varphi)$ then $C(1+\cos\varphi,0)$, let $M=tB+(1-t)O$ $=tB$ $=(t+t\cos\varphi,t\sin\varphi)$,
$CM\perp OB\Leftrightarrow$ $$ (t+t\cos\varphi-(1+\cos\varphi),t\sin\varphi).(1+\cos\varphi,\sin\varphi)=0$$ $$-2 \cos^2\left(\frac{\varphi}{2}\right) (\cos(\varphi) - 2 t + 1)=0$$ $$t=\frac{\cos(\varphi) + 1}{2}$$ $$M=t(1+\cos\varphi,\sin\varphi)= \frac{\cos(\varphi) + 1}{2}(1+\cos\varphi,\sin\varphi)$$
It's only left to scale up by $\frac{a}{2}$ back.
Update:
While $\varphi=\angle BOX_+$, $\angle BOA=\frac{1}{2}\angle BOX_+$, then $\angle BOX_+=2\angle BOA=2\theta$ and we're moving to polar coordinates. Then $r^2(\theta)=4\cos^6\theta$ and thus $r=2\cos^3\theta$ -- it's a Folium (case $\frac{b}{a}=4$).