Question

The radius of circle whose centre touches the line $x+2y+2z=15$ and sphere $x^2+y^2+z^2-2y-4z=11$ is

• A. $2$
• B. $\sqrt{7}$
• C. $3$
• D. $\sqrt{5}$

Answer 1

Answer: (B)

$\sqrt{7}$

Given equation of sphere is
$x^2+y^2+z^2-2y+4z=11$
Centre of sphere$=(0,1,2)$
and radius at sphere $=4$
Let centre of circle be $(\alpha ,\beta ,\gamma )$
The direction ratio's of line joining from centre of sphere to the centre of circle is $(\alpha -0,\beta -1,\gamma -2)$ or $(\alpha ,\beta -1,\gamma -2)$
But this line is normal at plane $x+2y+2z=15$
$\therefore \frac{\alpha }{1}=\frac{\beta -1}{2}=\frac{\gamma -2}{2}=k$
$\Rightarrow \alpha =k,\beta =2k+1,\gamma =2k+2$
$\because$ Centre of circle lies on $x+2y+2z=15$
$\therefore k+2(2k+1)+2(2k+2)=15\Rightarrow k=1$
So, centre of circle $=(1,3,4)$
Therefore, radius of circle
$\sqrt{{\left(\text{Radius of sphere}\right)}^2-{\left(\text{Length of the line joining centre of the circle and sphere}\right)}^2}$
$=\sqrt{{\left(4\right)}^2-[{(1-0)}^2+{(3-1)}^2+{(4-2)}^2]}=\sqrt{7}$