Question

Find an equation of the sphere with center (-3, 2 , 5) and radius 4. What is the intersection of this sphere with the yz-plane.

Answer 1

Answer:

The equation of the sphere with center (-3, 2 , 5) and radius 4 is $(x+3)^{2} +(y-2)^{2} + (z-5)^{2} = 16$

The intersection of the sphere with the yz- plane gave the equation $(y-2)^{2} + (z-5)^{2} = 7$ which is a 2D- circle with center (0,2,5) and radius $\sqrt{7}$.

Step-by-step explanation:

The equation of a sphere of radius r, with center (a,b,c) is given by

$(x-a)^{2} +(y-b)^{2} + (z-c)^{2} = r^{2}$

where, $x,$ $y,$ and $z$ are the coordinates of the points on the surface of the sphere.

Hence, the equation of the sphere with center,  (-3, 2 , 5) and radius 4 becomes

$(x-a)^{2} +(y-b)^{2} + (z-c)^{2} = r^{2}$

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

Then,

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

This is the equation of the sphere with center (-3, 2 , 5) and radius 4,

Now, for the intersection of this sphere with the yz- plane,

The $yz -$plane is where $x = 0$, then we set $x = 0$

Them the equation $(x+3)^{2} +(y-2)^{2} + (z-5)^{2} = 16$ becomes

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

$(3)^{2} +(y-2)^{2} + (z-5)^{2} = 16\\9 +(y-2)^{2} + (z-5)^{2} = 16\\(y-2)^{2} + (z-5)^{2} = 16 - 9\\(y-2)^{2} + (z-5)^{2} = 7$

This equation is the equation of a 2D- circle with center (0,2,5) and radius $\sqrt{7}$

This is the part of the sphere that intersects with the yz-plane.