Question

The radius of the sphere
$x^2+y^2+z^2=49,2x+3y-z-5\sqrt{14}=0$ is

• A. $\sqrt{6}$
• B. $2\sqrt{6}$
• C. $4\sqrt{6}$
• D. $6\sqrt{6}$

Answer 1

Answer: (B)

The sphere $x^2+y^2+z^2=49$
has centre at the origin (0, 0, 0) and radius 7.
Disance of the plane
$2x+3y-z-5-\sqrt{14}=0$
from the origin
$=\frac{|2\left(0\right)+3\left(0\right)-\left(0\right)-5\sqrt{14}|}{\sqrt{2^2+3^2+{\left(-1\right)}^2}}$
$=\frac{|-5\sqrt{14}|}{\sqrt{14}}=\frac{-5\sqrt{14}}{\sqrt{14}}=5$
Thus in Figure; $OP=7,ON=5$
$N{P}^2=O{P}^2-O{N}^2={\left(7\right)}^2-{\left(5\right)}^2$
$=49-25=24<br/>\therefore NP=2\sqrt{6}$ Hence the radius of the circle
$=NP=2\sqrt{6}$