Question

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

• A. $\sqrt{6}$
• B. $2\sqrt{6}$
• C. $4\sqrt{6}$
• D. none of these

Answer 1

The correct option is B $2\sqrt{6}$

The sphere $x^2+y^2+z^2=49$ has center at the origin $(0,0,0)$ and radius $7$.

Distance of the plane $2x+3y-z-5\sqrt{14}=0$ from the origin

$=\frac{|2\left(0\right)+3\left(0\right)-\left(0\right)-5\sqrt{14}|}{\sqrt{2^2+3^3+{(-1)}^2}}=\frac{|-5\sqrt{14}|}{\sqrt{14}}=\frac{5\sqrt{14}}{\sqrt{14}}=5$

${NP}^2={OP}^2-{ON}^2=7^2-5^2=49-25=24$

$\therefore NP=2\sqrt{6}$