Respuesta :
Answer:
Equation: [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]
Intersection: [tex](y-2)^2+(z-5)^2=7[/tex]
Step-by-step explanation:
We are asked to write an equation of the sphere with center center [tex](-3,2,5)[/tex] and radius 4.
We know that equation of a sphere with radius 'r' and center at [tex](h,k,l)[/tex] is in form:
[tex](x-h)^2+(y-k)^2+(z-l)^2=r^2[/tex]
Since center of the given sphere is at point [tex](-3,2,5)[/tex], so we will substitute [tex]h=-3[/tex], [tex]k=2[/tex], [tex]l=5[/tex] and [tex]r=4[/tex] in above equation as:
[tex](x-(-3))^2+(y-2)^2+(z-5)^2=4^2[/tex]
[tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]
Therefore, our required equation would be [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex].
To find the intersection of our sphere with the yz-plane, we will substitute [tex]x=0[/tex] in our equation as:
[tex](0+3)^2+(y-2)^2+(z-5)^2=16[/tex]
[tex]9+(y-2)^2+(z-5)^2=16[/tex]
[tex]9-9+(y-2)^2+(z-5)^2=16-9[/tex]
[tex](y-2)^2+(z-5)^2=7[/tex]
Therefore, the intersection of the given sphere with the yz-plane would be [tex](y-2)^2+(z-5)^2=7[/tex].
