1. ## write the equation

of a sphere witha center (1,4,-3) and tangent to the xz plane

2. Originally Posted by tiga killa
of a sphere witha center (1,4,-3) and tangent to the xz plane
$\displaystyle (x-1)^2 + (y-4)^2 + (z+3)^2 = r^2$

r is the distance from the point (1,4,-3) to the xz plane ... which I leave for you to figure out.

3. Hello, tiga killa!

Write the equaton of a sphere with center (1,4,-3) and tangent to the xz-plane
We have the center . . . we just need the radius, right?

Can you visualize the sphere?
. . If it is tangent to the xz-plane, the radius is 4.

Therefore: .$\displaystyle (x-1)^2 + (y-4)^2 + (x+3)^2 \:=\:16$

4. Hello,

You have just to find the radius of this sphere.

It corresponds of the abscissa since it has to be tangent to $\displaystyle (O_z)$.

Hence $\displaystyle \color{red}R=1$ and thus the equation of the sphere is $\displaystyle \color{red}(x-1)^2+(y-4)^2+(z+3)^2=1$

Edit : What I called $\displaystyle (O_z)$ is the plan $\displaystyle (Oyz)$, maybe the sphere is tangent to $\displaystyle (Oxz)$ ?