Question

The length of the intercept made by the sphere $x^2+y^2+z^2-4x+6y+8z+4=0$ on ${}^{\prime }{z}^{\prime }$ axis is

• A. $4\sqrt{3}$
• B. $2\sqrt{13}$
• C. $2\sqrt{15}$
• D. $2\sqrt{18}$

Answer 1

The correct option is A $4\sqrt{3}$

For $z$-intercept put $x=0,y=0$
$\Rightarrow 0^2+0^2+z^2-4\left(0\right)+6\left(0\right)+8z+4=0$
$\Rightarrow z^2+8z+4=0$
$\Rightarrow z=-4\pm 2\sqrt{3}$
Hence, length of the intercept is $=-4+2\sqrt{3}-(-4-2\sqrt{3})=4\sqrt{3}$