Question

The radius of the sphere
\[ x^2+y^2+z^2=12x+4y+3z \] is

• A. \(13/2\)
• B. \(13\)
• C. \(26\)
• D. \(52\)

Hint:

Sphere: \(x^2+y^2+z^2+2ux+2vy+2wz+d=0\). Radius \(=\sqrt{u^2+v^2+w^2-d}\).

Answer 1

The Correct Option is A

Step 1: Rewrite in standard sphere form.
\[ x^2-12x+y^2-4y+z^2-3z=0 \]
Step 2: Complete squares.
\[ (x-6)^2-36+(y-2)^2-4+\left(z-\frac{3}{2}\right)^2-\frac{9}{4}=0 \]
Step 3: Collect constants.
\[ (x-6)^2+(y-2)^2+\left(z-\frac{3}{2}\right)^2 =36+4+\frac{9}{4} \]
\[ =40+\frac{9}{4}=\frac{160+9}{4}=\frac{169}{4} \]
Step 4: Radius is square root of RHS.
\[ r=\sqrt{\frac{169}{4}}=\frac{13}{2} \]
Final Answer:
\[ \boxed{\frac{13}{2}} \]