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Chapter 12: Problem 13

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

Short Answer

Expert verified
The equation of the sphere is \((x + 3)^2 + (y - 2)^2 + (z - 5)^2 = 16\), and the intersection with the \(yz\)-plane is a circle \((y - 2)^2 + (z - 5)^2 = 7\).

Step by step solution

01

Write the General Equation of a Sphere

The general equation of a sphere with center \((h, k, l)\) and radius \(r\) is given by the equation \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\). In this problem, the center is \((-3, 2, 5)\) and the radius is 4.
02

Substitute the Values into the Sphere Equation

Substitute \(h = -3\), \(k = 2\), \(l = 5\), and \(r = 4\) into the general equation. This gives:\((x + 3)^2 + (y - 2)^2 + (z - 5)^2 = 4^2\).
03

Simplify the Radius Term

Calculate \(4^2\), which is 16. Thus, the equation of the sphere is:\((x + 3)^2 + (y - 2)^2 + (z - 5)^2 = 16\).
04

Identify the Intersection with the yz-plane

The intersection of the sphere with the \(yz\)-plane occurs when \(x = 0\). Substitute \(x = 0\) into the sphere's equation:\((0 + 3)^2 + (y - 2)^2 + (z - 5)^2 = 16\).
05

Simplify the Equation on the yz-plane

Simplify the equation by substituting \(3^2 = 9\): \(9 + (y - 2)^2 + (z - 5)^2 = 16\).Subtract 9 from both sides to get:\((y - 2)^2 + (z - 5)^2 = 7\).This represents a circle in the \(yz\)-plane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intersection with Plane
When we talk about the intersection of a sphere with a plane, it means we are looking for a set of points that both the sphere and the plane have in common. In the context of the given problem, our plane of interest is the **yz-plane**.

The yz-plane is the plane where the x-coordinate is always zero. This simplifies our calculations since we can set **x = 0** in the sphere鈥檚 equation.

  • Original sphere equation: \((x + 3)^2 + (y - 2)^2 + (z - 5)^2 = 16\)
  • Substitute **x = 0**: \((0 + 3)^2 + (y - 2)^2 + (z - 5)^2 = 16\)
  • Simplifies to: \(9 + (y - 2)^2 + (z - 5)^2 = 16\)


By subtracting 9 from both sides, we find:\[(y - 2)^2 + (z - 5)^2 = 7\]This equation represents the intersection of the sphere and the yz-plane, which forms a circle.
General Equation of a Sphere
The general equation of a sphere helps to define its shape in a three-dimensional space. It is derived from the center-radius form of the equation.

The standard form is:\[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]Here,
  • \( (h, k, l) \) - the coordinates of the center of the sphere
  • \( r \) - the radius of the sphere


This formula essentially states that any point \((x, y, z)\) on the surface of the sphere is a constant distance, called the radius, from its center \((h, k, l)\).

For example, the problem gives us a sphere with center \((-3, 2, 5)\) and radius \(4\). Substituting these values into the general equation, we get:\[(x + 3)^2 + (y - 2)^2 + (z - 5)^2 = 16\]
Circle in the YZ-Plane
Understanding the concept of a circle in the yz-plane is crucial when working with intersections. To visualize this, remember that when a sphere intersects the yz-plane, the x-value is set to zero. This turns the three-dimensional intersection on the plane into a 2-dimensional circle.In our practice problem, after substituting \(x = 0\) into the sphere's equation, we simplify it to:\[(y - 2)^2 + (z - 5)^2 = 7\]Let鈥檚 break this down:
  • The expression \((y - 2)\) means the circle is centered vertically at \(y = 2\).
  • Similarly, \((z - 5)\) indicates a horizontal center at \(z = 5\).
  • The number \(7\) represents the square of the circle's radius. Thus, the radius is \(\sqrt{7}\).


This equation describes a circle with center \((2,5)\) on the **yz-plane**, with a radius of \(\sqrt{7}\). Such visual and numerical insights enrich understanding of spatial relationships and geometric transformations.

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Most popular questions from this chapter

(a) Find the point at which the given lines intersect: \(\mathbf{r}=\langle 1,1,0\rangle+ t\langle 1,-1,2\rangle\) \(\mathbf{r}=\langle 2,0,2\rangle+ s\langle- 1,1,0\rangle\) (b) Find an equation of the plane that contains these lines.

Find the point at which the line intersects the given plane. $$ x=t-1, \quad y=1+2 t, \quad z=3-t ; \quad 3 x-y+2 z=5 $$

Find a vector equation and parametric equations for the line. The line through the point \((0,14,-10)\) and parallel to the line \(x=-1+2 t, y=6-3 t, z=3+9 t\)

Find the point at which the line intersects the given plane. $$ 5 x=y / 2=z+2 ; \quad 10 x-7 y+3 z+24=0 $$

\(35-36\) Find the volume of the parallelepiped with adjacent edges \(P Q, P R,\) and \(P S .\) $$ P(-2,1,0), \quad Q(2,3,2), \quad R(1,4,-1), \quad S(3,6,1) $$
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