Locus problem, circle

Question

A variable line moves in such way that the product of the perpendiculars form $(a,0)$ and $(0,0)$ is equal to $k^2$.The locus of the foot of the perpendicular from $(0,0)$ upon the variable line is a circle, the square of whose radius is? (Given : $|a| < 2|k|$)

Answer 1

Let $O=(0,0)$, $A=(a,0)$, $B=(X,Y)$ be the foot for $O$ and $C$ be the foot for $A$.

The equation of the dynamic line is $$\fbox{$X x+Y y = X^2+Y^2$}$$

Now,

\begin{align} OB \times AC &= k^2 \\ \left| \frac{-X^2-Y^2}{\sqrt{X^2+Y^2}} \times \frac{aX-X^2-Y^2}{\sqrt{X^2+Y^2}} \right| &= k^2 \\ X^2+Y^2-aX &= \pm k^2 \\ \left( X-\frac{a}{2} \right)^2+Y^2 &= \frac{a^2}{4} \pm k^2 \end{align}

Therefore,

$$\fbox{$r = \sqrt{\frac{a^2}{4} \pm k^2}$}$$

I'd like to show the geometrical interpretation here.

• Interestingly, $C$ also falls on the circle.

• For $X^2+Y^2-aX = k^2$ where $a,k\ne 0$

  

  Considering $OA$ as hypotenuse,

  $$(AC-OB)^2+BC^2=a^2$$

  Considering diameter $BB'$ or $CC'$ as hypotenuse,

  $$(AC+OB)^2+BC^2=4r^2$$

  Eliminating $BC$ gives $$\fbox{$AC\times OB =r^2-\frac{a^2}{4}$}$$

• For $X^2+Y^2-aX = -k^2$ where $a^2>4k^2$

  

  Considering $OA$ as hypotenuse,

  $$(AC+OB)^2+BC^2=a^2$$

  Considering diameter $BB'$ or $CC'$ as hypotenuse,

  $$(AC-OB)^2+BC^2=4r^2$$

  Eliminating $BC$ gives $$\fbox{$AC\times OB =\frac{a^2}{4}-r^2$}$$