Question

[melusa]

Tìm tọa độ tâm và bán kính mặt cầu sau đây:

a) x2+y2+z2-8x+2y+1=0

b) 3x2+3y2+3z2+6x-3y+15z-2=0

c) 9x2+9y2+9z2-6x+18y+1=0

 

Answer 1

[Hoàng Tử Hà]

a/ \(x^2-2.4x+16+y^2+2y+1+z^2=16\Leftrightarrow\left(x-4\right)^2+\left(y+1\right)^2+z^2=16\)

\(\Rightarrow\left\{{}\begin{matrix}I\left(4;-1;0\right)\\R=\sqrt{16}=4\end{matrix}\right.\)

b/ \(x^2+y^2+z^2+2x-y+5z-\dfrac{2}{3}=0\Leftrightarrow x^2+2x+1+y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+z^2+2.\dfrac{5}{2}z+\dfrac{25}{4}=\dfrac{2}{3}+1+\dfrac{1}{4}+\dfrac{25}{4}\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{2}\right)^2+\left(z+\dfrac{5}{2}\right)^2=\dfrac{49}{6}\) \(\Rightarrow\left\{{}\begin{matrix}I\left(-1;\dfrac{1}{2};-\dfrac{5}{2}\right)\\R=\dfrac{7}{\sqrt{6}}\end{matrix}\right.\)

P/s: câu c bạn tự làm nốt ạ!