Question

The center and radius of the sphere \(x^2+y^2+z^2-3x-4z+1=0\) are

• A. \(\left(-\frac{3}{2},0,-2\right),\frac{\sqrt{21}}{2}\)
• B. \(\left(\frac{3}{2},0,2\right),\sqrt{21}\)
• C. \(\left(\frac{3}{2},0,2\right),\frac{\sqrt{21}}{2}\)
• D. \(\left(-\frac{3}{2},0,2\right),\frac{21}{2}\)

Hint:

In \(x^2+y^2+z^2+2ux+2vy+2wz+d=0\), center is \((-u,-v,-w)\) and radius is \(\sqrt{u^2+v^2+w^2-d}\).

Answer 1

The Correct Option is C

Step 1: Compare with standard equation of sphere.
Standard form:
\[ x^2+y^2+z^2+2ux+2vy+2wz+d=0 \]
Center: \((-u,-v,-w)\)
Radius: \(\sqrt{u^2+v^2+w^2-d}\)
Step 2: Identify coefficients.
Given:
\[ x^2+y^2+z^2-3x-4z+1=0 \]
So:
\[ 2u=-3 \Rightarrow u=-\frac{3}{2} \]
\[ 2v=0 \Rightarrow v=0 \]
\[ 2w=-4 \Rightarrow w=-2 \]
\[ d=1 \]
Step 3: Find center.
\[ (-u,-v,-w)=\left(\frac{3}{2},0,2\right) \]
Step 4: Find radius.
\[ R=\sqrt{u^2+v^2+w^2-d} \]
\[ =\sqrt{\left(\frac{3}{2}\right)^2 + 0^2 + (2)^2 - 1} =\sqrt{\frac{9}{4}+4-1} =\sqrt{\frac{9}{4}+3} =\sqrt{\frac{21}{4}} =\frac{\sqrt{21}}{2} \]
Final Answer:
\[ \boxed{\left(\frac{3}{2},0,2\right),\frac{\sqrt{21}}{2}} \]