The radius of the sphere x2+y2+z2=x+2y+3z is
142
7
72
Solve for the radius of the sphere
x2+y2+z2=x+2y+3z
The equation of sphere is x2+y2+z2=x+2y+3z
Compared with the standard form,x2+y2+z2+2ux+2vy+2wz+d=0
The radius is given as, r=u2+v2+w2-d
So, the center is 12,1,32
The radius is given as,
=122+12+322-0=14+1+94=144=142
Hence, option(A) is the correct answer,
If the radius of the sphere x2+y2+z2−2x−4y−6z = 0 is r, then find the value of r2
Find the centre of the sphere x2+y2+z2+2z−x=0