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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2019 · Mathematics · Three Dimensional Geometry

What is the radius of the sphere x^2+y^2+z^2-6 x+8 y-10 z+1=0 ?

What is the radius of the sphere x2+y2+z2−6x+8y−10z+1=0x^2+y^2+z^2-6 x+8 y-10 z+1=0 ?

Options

  1. A5
  2. B2
  3. C7
  4. D3

Correct answer

C. 7

Step-by-step solution

The equation of sphere is x2+y2+z2−6x+8y−10z+1=0x^2+y^2+z^2-6 x+8 y-10 z+1=0 Comparing the equation with general form of sphere, x2+y2+z2+2ux+2vy+2wz−d=0x^2+y^2+z^2+2 u x+2 v y+2 w z-d=0 we get, =−62=−3,v=82=4,w=−102=−5,d=1Radius =u2+v2+w2−d=(−3)2+(4)2+(−5)2−1=9+16+25−1=49=7\begin{aligned} & \text{ we get, }=\frac{-6}{2}=-3, v=\frac{8}{2}=4, w=\frac{-10}{2}=-5, d=1 \\ & \text{Radius }=\sqrt{u^2+v^2+w^2-d} \\ & =\sqrt{(-3)^2+(4)^2+(-5)^2-1} \\ & =\sqrt{9+16+25-1} \\ & =\sqrt{49}=7\end{aligned}$$
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