Solveeit Logo
Login

Question

Mathematics Question on Circle

The center and radius of the sphere x2+y2+z2+3x−4z+1=0x^2 + y^2 + z^2 + 3x - 4z + 1 = 0 are

A

(−32,0,−2);212\left(-\frac{3}{2},0,-2\right);\frac{\sqrt{21}}{2}

B

(32,0,2);21\left(\frac{3}{2},0,2\right);\sqrt{21}

C

(−32,0,2);212\left(-\frac{3}{2},0,2\right);\frac{\sqrt{21}}{2}

D

(−32,0,2);212\left(-\frac{3}{2},0,2\right);\frac{21}{2}

Answer

(−32,0,2);212\left(-\frac{3}{2},0,2\right);\frac{\sqrt{21}}{2}

Explanation

Solution

Since the centre and radius of the sphere x2^2 + y2^2 + z2^2 + 2 ux + 2vy + 2wz + d = 0 are (-u, -v, - w) and u2+v2+w2−d\sqrt{u^{2} +v^{2} +w^{2} -d} respectively. So, for the sphere x2^2 + y2^2 + z2^2 + 3x - 4z + 1 = 0, Centre ≡(−32,0,2),\equiv\left(-\frac{3}{2},0,2\right), and Radius =(−32)2+02+(2)2−1\sqrt{\left(-\frac{3}{2}\right)^{2} +0^{2} +\left(2\right)^{2} -1} =94+4−1=212\quad\quad=\sqrt{\frac{9}{4}+4-1}=\frac{\sqrt{21}}{2}