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. The intersection of this sphere with the xz-plane results in a circle, as the value of y is fixed at 0 in the xz-plane. Therefore, the intersection can be expressed as (x - 1)² + (0 + 4)² + (z - 3)² = 25, simplifying to (x - 1)² + (z - 3)² = 9, which represents a circle with radius 3 centered at (1, 0, 3).

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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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