Question

The radius of the sphere $x^2+y^2+z^2=x+2y+3z$ is

• A. $\frac{\sqrt{14}}{2}$
• B. $\sqrt{7}$
• C. $\frac{7}{2}$
• D. $\frac{\sqrt{7}}{2}$

Answer 1

The correct option is A

$\frac{\sqrt{14}}{2}$

Solve for the radius of the sphere

$x^2+y^2+z^2=x+2y+3z$

The equation of sphere is $x^2+y^2+z^2=x+2y+3z$

Compared with the standard form,$x^2+y^2+z^2+2ux+2vy+2wz+d=0$

The radius is given as, $r=\sqrt{u^2+v^2+w^2-d}$

So, the center is $\left(\frac{1}{2},1,\frac{3}{2}\right)$

The radius is given as,

$$\begin{gathered} =\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(1\right)}^2+{\left(\frac{3}{2}\right)}^2-0} \\ =\sqrt{\frac{1}{4}+1+\frac{9}{4}} \\ =\sqrt{\frac{14}{4}} \\ =\frac{\sqrt{14}}{2} \end{gathered}$$

Hence, option(A) is the correct answer,