# Math Help - Spheres

1. ## Spheres

I'm very sory for asking another question, but I'm just preparing a final exam, which means that i solve something like 20 Problems a day, so I get confused from time to time

The Sphere S intersects the plane P in a circle. Find the radius of the circle. Given are:
Sphere S: x^2+y^2+z^2-2y-22z-103=0
Point C, centre of the circle of intersection: 6/7/8
and the plane P 2x+2y-z-18=0

My idea was first to find one direction vector of the plane, so for instance (1,-1,0) (one vector which lies on the plane)
Then I invented the line l, which goes from C farther on the plane an eventually hits the spehere.
l= (6,7,8)+t(1,-1,0)
To find where it hits the plane, I inserted the line in the equation of the sphere
(1+t)^2+(7-t)^2 +8^2-2(7-t)-22(8)-103=0
As result, I get t^2=(70)^1/2
This value inserted for t and than substracted the beginning point C, the radius of my circle gets (140)^1/2 which is the wrong solution. The result must be 12 (so (144)^1/2).
As many times as I perform the calculation, I get the wrong solution, where is the flaw of my approach??

2. Hello Schdero,

Try to find any point on the intersection between the plane and the sphere. For example, take x = 0 and replace in the equation of the plane and in the equation of the sphere. You get a system of 2 equations with two unknowns. Solve for y and z. If you did not succeed, try to take y = 0 and repeat. Now the radius is simply the distance between C and that point of intersection.

Hope this helps

3. Originally Posted by Schdero
I'm very sory for asking another question, but I'm just preparing a final exam, which means that i solve something like 20 Problems a day, so I get confused from time to time

The Sphere S intersects the plane P in a circle. Find the radius of the circle. Given are:
Sphere S: x^2+y^2+z^2-2y-22z-103=0
Point C, centre of the circle of intersection: 6/7/8
and the plane P 2x+2y-z-18=0

My idea was first to find one direction vector of the plane, so for instance (1,-1,0) (one vector which lies on the plane)
No, I think you are headed in the wrong direction: the problem is much simpler than you think. First, determine the center and radius of the sphere by completing the square. I get $S:\; (x-0)^2+(y-1)^2+(z-11)^2=15^2$. So the center of the sphere is $M(0/1/11)$ and its radius is $r=15$.
Since the center $C(6/7/9)$ of the circle of intersection of sphere and plane is given, the radius $\rho$ of the circle of intersection is one leg of a right triangle with hypotenuse $r=15$ and the other leg $|MC|$. Thus, $\rho=\sqrt{r^2-|MC|^2}=\sqrt{15^2-81}=12$.

4. Thank you, your solution is much simpler!

Schöne obe no ;-) (hope your not only swiss, but swiss-german^^)

5. Originally Posted by Schdero
I'm very sory for asking another question, but I'm just preparing a final exam, which means that i solve something like 20 Problems a day, so I get confused from time to time

The Sphere S intersects the plane P in a circle. Find the radius of the circle. Given are:
Sphere S: x^2+y^2+z^2-2y-22z-103=0
Point C, centre of the circle of intersection: 6/7/8
and the plane P 2x+2y-z-18=0

My idea was first to find one direction vector of the plane, so for instance (1,-1,0) (one vector which lies on the plane)
Then I invented the line l, which goes from C farther on the plane an eventually hits the spehere.
l= (6,7,8)+t(1,-1,0)
To find where it hits the plane, I inserted the line in the equation of the sphere
$({\color{red}1}+t)^2+(7-t)^2 +8^2-2(7-t)-22(8)-103=0$
As result, I get t^2=(70)^1/2
The above equation is wrong, it should be
$({\color{red}6}+t)^2+(7-t)^2 +8^2-2(7-t)-22(8)-103=0$
With this modification you get that $t_{1,2}=\pm \sqrt{72}=\pm 6\sqrt{2}$

This value inserted for t and than substracted the beginning point C, the radius of my circle gets (140)^1/2 which is the wrong solution. The result must be 12 (so (144)^1/2).
As many times as I perform the calculation, I get the wrong solution, where is the flaw of my approach??
Your approach is ok (although too complicated for my taste): it is only your calculation that was wrong.

6. I found that out too. Im sorry about that, i calculated it four times and got (70)^1/2=t every single time.
Thanks anyway
schdero