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

the radius of the sphere 3x2 +3y2 +3z2 -8x +4y +8z -15 = 0
 

Grade:12th pass

1 Answers

jagdish singh singh
173 Points
8 years ago
\hspace{-0.6 cm}$ Given $\bf{3x^2 +3y^2 +3z^2 -8x +4y +8z -15 = 0}$\\\\So $\bf{x^2+y^2+z^2-\frac{8}{3}x+\frac{4}{3}y+\frac{8}{3}z-5=0}$\\\\Now general eqn. of Sphere $\bf{x^2+y^2+z^2+2fx+2gy+2hx+c=0}$\\\\So we get $\bf{f=-\frac{8}{3}\;\;,g=-\frac{2}{3}\;\;,h=\frac{4}{3}\;.c=-5}$\\\\So center $\bf{(-f,-g,-h) = \left(\frac{8}{3}\;,\frac{2}{3}\;,-\frac{4}{3}\right)}.$ and $\bf{r=\sqrt{f^2+g^2+h^2-c}.}$

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