MHB S6.12.11 Find an equation of the sphere

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The equation of the sphere with center (1, -4, 3) and radius 5 is derived as (x - 1)² + (y + 4)² + (z - 3)² = 25. For the intersection of this sphere with the xz-plane, where y = 0, the equation simplifies to (x - 1)² + 16 + (z - 3)² = 25. This leads to the equation (x - 1)² + (z - 3)² = 9, representing a circle in the xz-plane with a radius of 3 centered at (1, 0, 3). The discussion emphasizes the relationship between the sphere's equation and its intersection with the coordinate planes. Understanding these geometric relationships is crucial for solving similar problems in three-dimensional space.
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$\tiny{s6.12.11}\\$
\begin{align}
&\textsf{(a) Find an equation of the sphere with center (1, -4, 3) and radius 5. }\\
&(x - 1)^2 +( y +4)^2 +(z - 3 )^2 = 5 ^2=25 \\
\\
&\textsf{(b) What is the intersection of this sphere with the
xz-plane?.}\\
&\textit{assume it is an equation of a circle for the intersection}\\
\end{align}
$\textit{}$
 
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For a sphere of radius $r$ and centered at $(a,b,c)$, we have:

$$(x-a)^2+(y-b)^2+(z-c)^2=r^2$$

What is the value of $y$ for all points in the $xz$-plane?
 
In the $xz$-plane, we have $y=0$...:D
 
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