Problem 43 - College Algebra and Calculus: An Applied Approach

Question

Find the sphere's center and radius. $$ x^{2}+y^{2}+z^{2}-2 x+6 y+8 z+1=0 $$

Step-by-Step Solution

Answer

The center of the sphere is at (1, -3, -4) and the radius is \(\sqrt{11}\)

Step 1: Convert the Given Equation into the General Form

Rearrange the given equation to align like terms and group them together: \( x^{2}-2x+y^{2}+6y+z^{2}+8z=-1 \). Complete the square for each of the variables x, y, z: \( (x-1)^{2}+(y+3)^{2}+(z+4)^{2}=11 \)

Step 2: Identify the Center

The equation in the general form is now \( (x-a)^{2}+(y-b)^{2}+(z-c)^{2}= r^{2} \). By comparing, we identify that \( a=1 \), \( b=-3 \), and \( c=-4 \). So, the center of the sphere is at (1, -3, -4)

Step 3: Identify the Radius

By comparing the converted equation with the general equation of the sphere, we can identify that \( r^{2}=11 \). By taking the square root of both sides, we get \( r=\sqrt{11} \)