A sphere sitting in the first octant and is tanget to xz and yz planes, has its center, C, in the plane z=5. The distance from the origin to C is sqrt(43).

I need to find an equation for this sphere.

What I know: equation of a sphere is $$(x-a)^2+(y-b)^2+(z-c)^2=r^2$$
And that the distance between two points in R^3 is $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)}$$

Since Im not given much info, what is my first step in finding an equation for this sphere? Without an x and y coordinate for C is it possible to use the distance formula for anything?

Question

A sphere sitting in the first octant and is tanget to xz and yz planes, has its center, C, in the plane z=5. The distance from the origin to C is sqrt(43).

I need to find an equation for this sphere.

What I know: equation of a sphere is $$(x-a)^2+(y-b)^2+(z-c)^2=r^2$$
And that the distance between two points in R^3 is $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)}$$

Since Im not given much info, what is my first step in finding an equation for this sphere? Without an x and y coordinate for C is it possible to use the distance formula for anything?

anonymous · Answer

Since I am given the distance from (0,0,0) to the center is sqrt(43) is it fair to say that  $$\sqrt{43}=\sqrt{(x_2-0)^2+(y_2-0)^2+(5-0)^2}$$
If so then $$\sqrt{43}=\sqrt{x^2+y^2+25}$$
and x and y = +/- 3.
Which would mean x and y = 3 because we are in the first octant right?
Any opinions?

anonymous · Answer

So this puts C at (3,3,5) which while I dont know if I am right, seems to hold up given what I know of the sphere, and using the distance formula with everything plugged in.

anonymous · Answer

Given a sphere with center (3,3,5) that is tangent to the xz and yz planes, I can say this sphere is radius 3, because the sphere must exactly touch the xz and yz planes, which are 3 units from the center... Does this sound right? Anybody?