Question

Find the equation of a sphere through the four points $(0,0,0),(-1,1,1),(1,-1,1),(1,1,-1)$ and determine its radius.

• A. $\frac{3\sqrt{3}}{2}$
• B. $\frac{3\sqrt{2}}{2}$
• C. $\frac{3}{2}$
• D. $\frac{5\sqrt{3}}{2}$

Answer 1

The correct option is A $\frac{3\sqrt{3}}{2}$

Since the sphere passes through origin, its equation is of the form
$x^2+y^2+z^2+2ux+2vy+2wz=0.$ ...(1)
Substituting the co - ordinates of the other given points, we get
$a^2+b^2+c^2-2ua+2vb+2wc=0.$ ...(2)
$a^2+b^2+c^2-2ua-2vb+2wc=0.$ ...(3)
$a^2+b^2+c^2+2ua+2vb-2wc=0.$ ...(4)
Adding (2) and (3), $2(a^2+b^2+c^2)=-4wc$
$2w=\frac{a^2+b^2+c}{-b}.$
Similarly, $2v=\frac{a^2+b^2+c}{-c}$ and $2v=\frac{a^2+b^2+c}{-a}$
Putting the values of u, v, w in (1), we get
$x^2+y^2+z^2-(a^2+b^2+c^2)(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})=0.$
The radius of the sphere is $\sqrt{(u^2+v^2+w^2)},d$ being zero
$R=\sqrt{\left\{\frac{{(a^2+b^2+c^2)}^2}{4}(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\right\}}$
$=\frac{a^2+b^2+c^2}{2}\sqrt{(a^{-2}+b^{-2}+c^{-2})}.$
Here $a=1,b=1,c=1$. Thus radius of the sphere is $R=\frac{3\sqrt{3}}{2}$