Practicing Success
The equation of the sphere which passes through the points (1, 0, 0), (0, 1, 0) and (0, 0, 1) and has its radius as small as possible is : |
3(x2 + y2 + z2) + 2(x + y + z) – 1= 0 3(x2 + y2 + z2) – 2(x + y + z) – 1= 0 (x2 + y2 + z2) – 2(x + y + z) – 1= 0 (x2 + y2 + z2) + 2(x + y + z) – 1= 0 |
3(x2 + y2 + z2) – 2(x + y + z) – 1= 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 $x^2+y^2+z^2-\frac{2}{3}(x+y+z)-\frac{1}{3}=0 \Rightarrow 3\left(x^2+y^2+z^2\right)-2(x+y+z)-1=0$ |