Question

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

• A. $2\sqrt{6}$
• B. $6\sqrt{2}$
• C. $\sqrt{14}$
• D. $5$

Answer 1

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

Centre of sphere $x^2+y^2+z^2=49$ is the origin $(0,0,0)$ and radius $r=7$
Let $p=$ length of $\perp$ er from $(0,0,0)$ to plane $2x+3y-z=5\sqrt{14}$
$=\frac{|2\times 03\times 0-0-5\sqrt{15}|}{\sqrt{14}}=5$
$\therefore R=\sqrt{r^2-p^2}=\sqrt{7^2-5^2}=2\sqrt{6}$
Hence choice (a) is correct answer.