Question

The sphere whose centre $=(\alpha ,\beta ,\gamma )$ and radius $=a,$ has the equation ${(x-\alpha )}^2+{(y-\beta )}^2+{(z-y)}^2=a^2$. Again, the intersection of a sphere by a plane is a circle. The distance of the centre of the sphere $x^2+y^2+z^2-2x-4y=0$ from the origin is

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

Answer 1

Answer: (B)

$\sqrt{5}$

Intersection of a sphere by a plane is circle equation of sphere $x^2+y^2+z^2-2x-4y=0$, centre $(1,2,0)$

now it is know that centre of the sphere $=(1,2,0)$

so distance b/w $(1,2,0)$ and origin $(0,0,0)$

is given by distance formula $=\sqrt{(1-0{)}^2+(2-0{)}^2+(0-0{)}^2}$

$=\sqrt{1+4+0}$

$=\sqrt{5}$

So, the correct option is $B\to \sqrt{5}$