The equation of the sphere which passes through the points (1, 0, 0),…

Let equation of sphere be given by

$x^2+y^2+z^2+2 u x+2 v y+2 w z+d=0$                 ……(1)

As sphere passes through points (1, 0, 0), (0, 1, 0) and (0, 0, 1). So we have

1 + 2u + d = 0, 1 + 2v + d = 0, 1 + 2w + d = 0

On solving u = v = w = $-\frac{1}{2}$ (d + 1)

If r is the radius of the sphere, then

$r=\sqrt{u^2+v^2+w^2-d}$

$r^2=\frac{3}{4}(d+1)^2-d=\mu$ (say)

for r to be minimum

$\frac{dµ}{dd}=0 \Rightarrow \frac{3}{4} . 2(d+1) - 1 = 0$  or  $d = -\frac{1}{3}$

Also, $\frac{d^2 \mu}{d^2}=\frac{3}{2}=$ positive at $d=-\frac{1}{3}$

Hence µ is minimum at d = $-\frac{1}{3}$

So, substituting value of d we have u = v = w = $-\frac{1}{3}$

∴ equation of the sphere

$x^2+y^2+z^2-\frac{2}{3}(x+y+z)-\frac{1}{3}=0 \Rightarrow 3\left(x^2+y^2+z^2\right)-2(x+y+z)-1=0$