Locus of the center of the circle of radius $a$,which always intersects coordinate axes

Question

If the axes are rectangular, show that the locus of the center of the circle of radius $a$,which always intersects coordinate axes is

$x\sqrt{a^2-y^2-z^2}+y\sqrt{a^2-z^2-x^2}+z\sqrt{a^2-x^2-y^2}=a^2$

Let the circle intersect the axes at $(x_1,0,0),(0,y_1,0),(0,0,z_1)$.

Let the center of the circle be $(x_0,y_0,z_0)$ and the radius of the circle be $a$.So,

$(x_0-x_1)^2+(y_0-0)^2+(z_0-0)^2=(x_0-0)^2+(y_0-y_1)^2+(z_0-0)^2=(x_0-0)^2+(y_0-0)^2+(z_0-z_1)^2=a^2..............(1)$

Also since the point $(x_0,y_0,z_0)$ lies on the plane $\dfrac{x}{x_1}+\dfrac{y}{y_1}+\dfrac{z}{z_1}=0$,

$\dfrac{x_0}{x_1}+\dfrac{y_0}{y_1}+\dfrac{z_0}{z_1}=0..................(2)$

I am stuck here. I cannot prove

$x_0\sqrt{a^2-y_0^2-z_0^2}+y_0\sqrt{a^2-z_0^2-x_0^2}+z_0\sqrt{a^2-x_0^2-y_0^2}=a^2$ from equations $(1),(2)$.

Please help.

Answer 1

Also since the point $(x_0,y_0,z_0)$ lies on the plane $\dfrac{x}{x_1}+\dfrac{y}{y_1}+\dfrac{z}{z_1}=0$,

$\dfrac{x_0}{x_1}+\dfrac{y_0}{y_1}+\dfrac{z_0}{z_1}=0..................(2)$

This is not correct. It should be $$\dfrac{x_0}{x_1}+\dfrac{y_0}{y_1}+\dfrac{z_0}{z_1}=\color{red}{1}\tag2$$

From $(1)$, subtracting $$x_0^2+(y_0-y_1)^2+z_0^2=a^2$$ from $$(x_0-x_1)^2+y_0^2+z_0^2=a^2\tag3$$ gives $$x_1^2-2x_0x_1+2y_0y_1-y_1^2=0\implies x_1=x_0\pm\sqrt{x_0^2-2y_0y_1+y_1^2}\tag4$$ Also, subtracting $$x_0^2+(y_0-y_1)^2+z_0^2=a^2$$ from $$x_0^2+y_0^2+(z_0-z_1)^2=a^2$$ gives $$-y_1^2+2y_0y_1-2z_0z_1+z_1^2=0\implies z_1=z_0\pm\sqrt{z_0^2-2y_0y_1+y_1^2}\tag5$$ Then, using $(3)(4)(5)$ $$y_1^2-2y_0y_1+x_0^2+y_0^2+z_0^2-a^2=0\implies y_1=y_0\pm\sqrt{a^2-x_0^2-z_0^2}$$ and so from $(4)(5)$

$$x_1=x_0\pm\sqrt{a^2-y_0^2-z_0^2},\quad z_1=z_0\pm\sqrt{a^2-x_0^2-y_0^2}$$

Finally, from $(2)$, $$\dfrac{x_0}{x_0\pm\sqrt{a^2-y_0^2-z_0^2}}+\dfrac{y_0}{y_0\pm\sqrt{a^2-x_0^2-z_0^2}}+\dfrac{z_0}{z_0\pm\sqrt{a^2-x_0^2-y_0^2}}=1$$ $$\dfrac{x_0\left(x_0\mp\sqrt{a^2-y_0^2-z_0^2}\right)}{x_0^2+y_0^2+z_0^2-a^2}+\dfrac{y_0\left(y_0\mp\sqrt{a^2-x_0^2-z_0^2}\right)}{x_0^2+y_0^2+z_0^2-a^2}+\dfrac{z_0\left(z_0\mp\sqrt{a^2-x_0^2-y_0^2}\right)}{x_0^2+y_0^2+z_0^2-a^2}=1$$

$$\small x_0\left(x_0\mp\sqrt{a^2-y_0^2-z_0^2}\right)+y_0\left(y_0\mp\sqrt{a^2-x_0^2-z_0^2}\right)+z_0\left(z_0\mp\sqrt{a^2-x_0^2-y_0^2}\right)=x_0^2+y_0^2+z_0^2-a^2$$

$$\small x_0^2\mp x_0\sqrt{a^2-y_0^2-z_0^2}+y_0^2\mp y_0\sqrt{a^2-x_0^2-z_0^2}+z_0^2\mp z_0\sqrt{a^2-x_0^2-y_0^2}=x_0^2+y_0^2+z_0^2-a^2$$ $$\pm x_0\sqrt{a^2-y_0^2-z_0^2}\pm y_0\sqrt{a^2-x_0^2-z_0^2}\pm z_0\sqrt{a^2-x_0^2-y_0^2}=a^2$$ as desired.

(I think that $$x_0\sqrt{a^2-y_0^2-z_0^2}+y_0\sqrt{a^2-z_0^2-x_0^2}+z_0\sqrt{a^2-x_0^2-y_0^2}=a^2$$ does not always hold. Take, for example, $x_0=-1, y_0=z_0=0, x_1=y_1=z_1=0, a=1$.)