Tim tâm \(I\) và bán kính \(R\) mặt cầu \(\left(S\right):3x^2+3y^2+3z^2-6x+8y+15z-3=0\).
\(I\left(3;-4;-\dfrac{15}{2}\right),R=\sqrt{3}\).\(I\left(1;-\dfrac{4}{3};-\dfrac{5}{2}\right),R=\dfrac{19}{6}\).\(I\left(1;\dfrac{4}{3};\dfrac{5}{2}\right),R=\dfrac{19}{6}\).\(I\left(1;\dfrac{4}{3};-\dfrac{5}{2}\right),R=\dfrac{3}{4}\).Hướng dẫn giải:Ta có:
\(3x^2+3y^2+3z^2-6x+8y+15z-3=0\)\(\Leftrightarrow x^2+y^2+z^2-2x+\dfrac{8}{3}y+5z-1=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+2.\dfrac{4}{3}y+\dfrac{16}{9}\right)+\left(z^2+2.\dfrac{5}{2}z+\dfrac{25}{4}\right)-1-\dfrac{16}{9}-\dfrac{25}{4}-1=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+\dfrac{4}{3}\right)^2+\left(z+\dfrac{5}{2}\right)^2=\dfrac{361}{36}\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+\dfrac{4}{3}\right)^2+\left(z+\dfrac{5}{2}\right)^2=\left(\dfrac{19}{6}\right)^2\)
=> Tâm mặt cầu là \(I\left(1;-\dfrac{4}{3};-\dfrac{5}{2}\right)\) và bán kính \(R=\dfrac{19}{6}\)
