write the equation

• May 5th 2009, 12:27 PM
tiga killa
write the equation
of a sphere witha center (1,4,-3) and tangent to the xz plane
• May 5th 2009, 01:34 PM
skeeter
Quote:

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.
• May 5th 2009, 01:38 PM
Soroban
Hello, tiga killa!

Quote:

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\$

• May 5th 2009, 01:43 PM
Infophile
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)\$ ?

:)