Spheres

Printable View

April 1st 2010, 11:43 PM
Schdero

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(Worried)

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??
April 2nd 2010, 01:18 AM
mohammadfawaz

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
April 2nd 2010, 09:53 AM
Failure

Quote:

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(Worried)

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, $\rho=\sqrt{r^2-|MC|^2}=\sqrt{15^2-81}=12$ .
April 3rd 2010, 12:13 PM
Schdero

Thank you, your solution is much simpler!

Schöne obe no ;-) (hope your not only swiss, but swiss-german^^)
April 3rd 2010, 08:25 PM
Failure

Quote:

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(Worried)

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 $t_{1,2}=\pm \sqrt{72}=\pm 6\sqrt{2}$

Quote:

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.
April 4th 2010, 01:03 AM
Schdero

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(Headbang)
schdero