GROUP 12 Sphere-Sphere Collision
Group $\cdot 4$ questions
Given the two spheres:
\[
S_{1}:(z-1)^{2}+(y-2)^{2}+(z-3)^{2}=9 \text { and } S_{2}:(z-5)^{2}+(y-5)^{2}+(z-3)^{2}=4
\]
Answer the 4 questions below
QUESION 121
Find First Sphere center and radius
Choose one $\cdot 5$ points
Find the center $\overrightarrow{c_{1}}$ and radius $R_{1}$ of sphere $S_{1}$
\[
\begin{array}{l}
\vec{c}_{1}=(1,2,3), R_{1}=3 \\
\vec{c}_{1}=(-1,-2,-3), R_{1}=3 \\
\overrightarrow{c_{1}=}(-1,-2,3), R_{1}=9
\end{array}
\]
Answer
\(\boxed{\vec{c}_{1}=(1,2,3), R_{1}=3}\) is the final answer.
Step 1 ：The equation of a sphere in 3D space is given by \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\), where \((a, b, c)\) is the center of the sphere and \(r\) is the radius. By comparing this with the given equation of the sphere \(S_1\), we can directly read off the coordinates of the center and the radius.
Step 2 ：The coefficients of \(S_1\) are \([1, 2, 3, 9]\).
Step 3 ：So, the center of the sphere \(S_{1}\) is \(\vec{c}_{1}=(1,2,3)\) and the radius \(R_{1}=3\).
Step 4 ：\(\boxed{\vec{c}_{1}=(1,2,3), R_{1}=3}\) is the final answer.
