Answer: The equation of the sphere with the center and radius

b) The intersection of this sphere with the y z-plane the x- co-ordinate
is zero(i.e., x = 0 )
Step-by-step explanation:
a) The equation of the sphere having center (h,k,l) and radius r is

Given center of the sphere (3, -9, 3) and radius 5

on simplification , we get solution


Final answer :-

b) The intersection of this sphere with the y z-plane the x- co-ordinate
is zero(i.e., x = 0 )

For a better understanding of the solution given here please find the attached file which has the relevant diagram.
To answer this question we will have to make use of <u>the Isosceles Triangle Theorem</u> which states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle. Thus, as a corollary we know that is EF is the angle bisector of the vertex angle ∠E, then, EF is the perpendicular bisector of the of the base DK.
Please follow the diagram of a complete understanding of the logic and the solution.
As EF is the angle bisector as given in the question, thus we will have:
.
Also, from the Theorem we know that KF will be half of DK and thus, KF will be:
centimeters.
Likewise, from the same theorem we have: 
the distance form X to Y is clearly -6 to 0 is 6 units, and 0 to 8 is 8 units, so 6 + 8 = 14 units.
now, for XZ and ZY we can simply use as stated, the distance formula to get those and then add them all to get the perimeter.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ X(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\qquad Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ XZ=\sqrt{[5-(-6)]^2+[8-2]^2}\implies XZ=\sqrt{(5+6)^2+(8-2)^2} \\\\\\ XZ=\sqrt{121+36}\implies \boxed{XZ=\sqrt{157}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20X%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B2%7D%29%5Cqquad%20Z%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B8%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20XZ%3D%5Csqrt%7B%5B5-%28-6%29%5D%5E2%2B%5B8-2%5D%5E2%7D%5Cimplies%20XZ%3D%5Csqrt%7B%285%2B6%29%5E2%2B%288-2%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20XZ%3D%5Csqrt%7B121%2B36%7D%5Cimplies%20%5Cboxed%7BXZ%3D%5Csqrt%7B157%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad Y(\stackrel{x_2}{8}~,~\stackrel{y_2}{2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ ZY=\sqrt{(8-5)^2+(2-8)^2}\implies ZY=\sqrt{9+36}\implies \boxed{ZY=\sqrt{45}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{perimeter}{14+\sqrt{157}+\sqrt{45}}\qquad \approx \qquad 33.2](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20Z%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B8%7D%29%5Cqquad%20Y%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B2%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20ZY%3D%5Csqrt%7B%288-5%29%5E2%2B%282-8%29%5E2%7D%5Cimplies%20ZY%3D%5Csqrt%7B9%2B36%7D%5Cimplies%20%5Cboxed%7BZY%3D%5Csqrt%7B45%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bperimeter%7D%7B14%2B%5Csqrt%7B157%7D%2B%5Csqrt%7B45%7D%7D%5Cqquad%20%5Capprox%20%5Cqquad%2033.2)
I don't know how many boys are in the classroom?
Your answer is correct. Good job.