Question

Get the radius of the circle that is the intersection of the sphere $x^2+y^2+z^2=49$ and the plane $2x+3y-z=5\sqrt{14}$.

Answer 1

$x^2+y^2+z^2=49$

Radius of the sphere$=7$ unit

it's centre is $(0,0,0)$

The intersecting plane is $2x+3y-z=5\sqrt{14}$

Distance of the plane from the centre of the sphere is

$=\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}$

$=\frac{2\times 0+3\times 0-0-5\sqrt{14}}{\sqrt{2^2+3^2+1^2}}=5$ units

In right angle triangle $OAB$

$AB$ is the radius of the circle

$AB=\sqrt{O{A}^2-O{B}^2}=\sqrt{7^2-5^2}=\sqrt{49-25}=2\sqrt{6}$ units.