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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.

User Kanghee
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1 Answer

5 votes

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
√(7).

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
√(7)

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

User Ruhul Amin
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7.0k points