Question

A line passes through the centre of a sphere whose radius is 5 and one of the intercept points is $(1,-2,2)$. If the equation of the line is
$\frac{x}{1}=\frac{y}{-2}=\frac{z}{2}$
,then the equation of the sphere can be

• A. $x^2+y^2+z^2-\frac{16}{3}x+\frac{32}{3}y-\frac{32}{3}z=-39$
• B. $x^2+y^2+z^2+\frac{4}{3}x-\frac{8}{3}y+\frac{8}{3}z-21=0$
• C. $x^2+y^2+z^2+\frac{16}{3}x-\frac{32}{3}y+\frac{32}{3}z=-39$
• D. $x^2+y^2+z^2-\frac{16}{3}x+\frac{32}{3}y-\frac{32}{3}z=39$

Answer 1

Answer: (A), (B)

The correct options are
A $x^2+y^2+z^2-\frac{16}{3}x+\frac{32}{3}y-\frac{32}{3}z=-39$
B $x^2+y^2+z^2+\frac{4}{3}x-\frac{8}{3}y+\frac{8}{3}z-21=0$

Since line passes through the centre,
$\frac{x}{1}=\frac{y}{-2}=\frac{z}{2}=1\mathrm{So},\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{line}\mathrm{can}\mathrm{be}\mathrm{represented}\mathrm{as}(t,-2t,2t).$
Let's assume that this point is the centre of the sphere, then according to the question, radius of sphere is 5. So,
$\sqrt{{\left(t-1\right)}^2+{\left(-2t+2\right)}^2+{\left(2t-2\right)}^2}=5\sqrt{{\left(t-1\right)}^2+4{\left(t-1\right)}^2+4{\left(t-1\right)}^2}=5\sqrt{9{\left(t-1\right)}^2}=53(t-1)=5,3(1-t)=5t=\frac{8}{3},\frac{-2}{3}$
So the coordinates of center of the sphere can be $\left(\frac{8}{3},-\frac{16}{3},\frac{16}{3}\right)$ or $\left(\frac{-2}{3},\frac{4}{3},\frac{-4}{3}\right)$.
The equation of the sphere in the first case is
${\left(x-\frac{8}{3}\right)}^2+{\left(y+\frac{16}{3}\right)}^2+{\left(z-\frac{16}{3}\right)}^2=5^2{x}^2+y^2+z^2-\frac{16}{3}x+\frac{32}{3}y-\frac{32}{3}z=-39$
The equation of the sphere in the second case is
${\left(x+\frac{2}{3}\right)}^2+{\left(y-\frac{4}{3}\right)}^2+{\left(z+\frac{4}{3}\right)}^2=5^2{x}^2+y^2+z^2+\frac{4}{3}x-\frac{8}{3}y+\frac{8}{3}z-21=0$