Question

Find an equation of the sphere with center s23, 2, 5d and radius 4. What is the intersection of this sphere with the yz-plane?

Answer 1

Answer:

Equation: $(x+3)^2+(y-2)^2+(z-5)^2=16$

Intersection: $(y-2)^2+(z-5)^2=7$

Step-by-step explanation:        

We are asked to write an equation of the sphere with center center $(-3,2,5)$ and radius 4.    

We know that equation of a sphere with radius 'r' and center at $(h,k,l)$ is in form:

$(x-h)^2+(y-k)^2+(z-l)^2=r^2$

Since center of the given sphere is at point $(-3,2,5)$, so we will substitute $h=-3$, $k=2$, $l=5$ and $r=4$ in above equation as:

$(x-(-3))^2+(y-2)^2+(z-5)^2=4^2$

$(x+3)^2+(y-2)^2+(z-5)^2=16$

Therefore, our required equation would be $(x+3)^2+(y-2)^2+(z-5)^2=16$.

To find the intersection of our sphere with the yz-plane, we will substitute $x=0$ in our equation as:

$(0+3)^2+(y-2)^2+(z-5)^2=16$

$9+(y-2)^2+(z-5)^2=16$

$9-9+(y-2)^2+(z-5)^2=16-9$

$(y-2)^2+(z-5)^2=7$

Therefore, the intersection of the given sphere with the yz-plane would be $(y-2)^2+(z-5)^2=7$.