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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2015 · Mathematics · Three Dimensional Geometry

The radius of the sphere 3 x^2+3 y^2+3 z^2-8 x+4 y+8 z-15=0 is

The radius of the sphere 3 x^2+3 y^2+3 z^2-8 x+4 y+8 z-15=0 \text{ is }

Options

  1. A2
  2. B3
  3. C4
  4. D5

Correct answer

B. 3

Step-by-step solution

3x2+3y2+3z2−8y+4y+8z−15=0⇒x2+y2+z2−83x+43y+83z−5=0⇒(x−43)2−169+(y+23)2−49+(z+43)2−169−5=0(x−43)2+(y+23)2+(z+43)2=(3)2\begin{aligned} & 3 x^2+3 y^2+3 z^2-8 y+4 y+8 z-15=0 \\ \Rightarrow & x^2+y^2+z^2-\frac{8}{3} x+\frac{4}{3} y+\frac{8}{3} z-5=0 \\ \Rightarrow & \left(x-\frac{4}{3}\right)^2-\frac{16}{9}+\left(y+\frac{2}{3}\right)^2-\frac{4}{9}+\left(z+\frac{4}{3}\right)^2-\frac{16}{9}-5=0 \\ & \left(x-\frac{4}{3}\right)^2+\left(y+\frac{2}{3}\right)^2+\left(z+\frac{4}{3}\right)^2=(3)^2 \end{aligned} So radius is 3.
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