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Animated Solution for Mathematics - Three Dimensional Geometry: The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0$ is cut by the plane $x + 2 y + 2 z + 7 = 0$ is

Q: What if the value inside the square root for $r$ turns out to be negative?

That is a great question! If $R^2 - d^2 R$. Geometrically, this implies the perpendicular distance from the center to the plane is greater than the radius of the sphere. In such a case, the plane is too far away to touch the sphere, so there is no intersection. In this problem, we found $R=5$ and $d=4$, so $25 - 16 = 9$, which is positive, confirming the intersection exists!

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Progressive Hints

Hint 1

Visualize the geometry: When a plane intersects a sphere, the cross-section is always a circle. Think about the right-angled triangle formed by the sphere's center, the center of the intersection circle, and any point on the circle's circumference. What are the lengths of the sides of this triangle?

Hint 2

Recall the standard form of a sphere equation:

x^{2} + y^{2} + z^{2} + 2 ux + 2 v y + 2 w z + d_{s p h er e} = 0

. How do you extract the center

O (- u, - v, - w)

and the radius

R = u^{2} + v^{2} + w^{2} - d_{s p h er e}

? Also, remember the formula for the perpendicular distance

d

from a point

(x_{1}, y_{1}, z_{1})

to a plane

a x + b y + cz + d_{pl an e} = 0

Hint 3

Start by identifying the sphere's center

O

and radius

R

from the given equation. Then, calculate the perpendicular distance

d

from

O

to the plane

x + 2 y + 2 z + 7 = 0

. Finally, use the Pythagorean relationship

r = R^{2} - d^{2}

to find the radius

r

of the intersection circle.

Visualized Solution

Visualizing the Sphere-Plane Intersection

The intersection of a sphere and a plane is always a circle.
Let $O$ be the center of the sphere and $R$ be its radius.
Let $C$ be the center of the intersection circle and $r$ be its radius.
The distance $O C = d$ is the perpendicular distance from $O$ to the plane.

The Geometry of the Cut

In the right-angled triangle $△ O C A$ , the angle at $C$ is $9 0^{\circ}$ .
By Pythagoras Theorem, we have: $O A^{2} = O C^{2} + A C^{2}$
This simplifies to the relation: $R^{2} = d^{2} + r^{2}$
Solving for the circle's radius: $r = R^{2} - d^{2}$

Finding the Center of the Sphere $O$

The general equation of a sphere is $x^{2} + y^{2} + z^{2} + 2 ux + 2 v y + 2 w z + d_{s p h er e} = 0$ .
Comparing with $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0$ :
$2 u = 2 ⟹ u = 1$
$2 v = - 2 ⟹ v = - 1$
$2 w = - 4 ⟹ w = - 2$
The center is $O (- u, - v, - w) = (- 1, 1, 2)$ .

Calculating the Sphere's Radius $R$

The radius of the sphere is given by: $R = u^{2} + v^{2} + w^{2} - d_{s p h er e}$
Substitute $u = 1$ , $v = - 1$ , $w = - 2$ , and $d_{s p h er e} = - 19$ :
$R = 1^{2} + (- 1)^{2} + (- 2)^{2} - (- 19)$
$R = 1 + 1 + 4 + 19 = 25 = 5$

Perpendicular Distance Formula

The perpendicular distance $d$ from a point $(x_{1}, y_{1}, z_{1})$ to a plane $a x + b y + cz + d_{pl an e} = 0$ is:
$d = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d _{pl an e} ∣}{a ^{2} + b ^{2} + c ^{2}}$
This formula gives the shortest distance from the sphere's center to the cutting plane.

Calculating the Distance $d$

Substitute the center $O (- 1, 1, 2)$ and the plane $x + 2 y + 2 z + 7 = 0$ into the formula:
$d = \frac{∣1 ( - 1 ) + 2 ( 1 ) + 2 ( 2 ) + 7∣}{1 ^{2} + 2 ^{2} + 2 ^{2}}$
$d = \frac{∣ - 1 + 2 + 4 + 7∣}{1 + 4 + 4} = \frac{12}{9} = \frac{12}{3} = 4$

Finding the Circle's Radius $r$

Now substitute $R = 5$ and $d = 4$ into the Pythagoras relation:
$r = R^{2} - d^{2}$
$r = 5^{2} - 4^{2} = 25 - 16$
$r = 9 = 3$

Summary of Sphere-Plane Intersections

The radius of the intersection circle is $r = 3$ .
If $d = 0$ , the plane passes through the center, forming a Great Circle ( $r = R$ ).
If $d = R$ , the plane is tangent to the sphere ( $r = 0$ ).
If $d > R$ , the plane does not intersect the sphere.

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The Sigma Insight: Equation of a Plane

Analyzing the Setup

Welcome, fellow traveler of the mathematical cosmos! Today, we are embarking on a journey into the heart of three-dimensional geometry.

Imagine a perfect sphere defined by the equation:

x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0

Our mission is to determine the radius of the circle formed when the plane

x + 2 y + 2 z + 7 = 0

slices through this sphere. This exploration relies on the fundamental symmetry of spatial relationships.

Phase 1

Unmasking the Sphere

To understand the sphere, we compare the given equation to the general form

x^{2} + y^{2} + z^{2} + 2 ux + 2 v y + 2 w z + d_{s p h er e} = 0

. By identifying the coefficients, we find

u = 1

v = - 1

, and

w = - 2

The center of the sphere,

O (- u, - v, - w)

, is located at

(- 1, 1, 2)

. This point serves as the heart of our sphere.

Next, we calculate the radius

R

using the formula

R = u^{2} + v^{2} + w^{2} - d_{s p h er e}

R = 1^{2} + (- 1)^{2} + (- 2)^{2} - (- 19)

R = 1 + 1 + 4 + 19 = 25 = 5

Phase 2

The Plane's Intrusion

The plane

x + 2 y + 2 z + 7 = 0

acts as our cutting tool. To find the radius of the intersection circle, we must first determine the perpendicular distance

d

from the center

O (- 1, 1, 2)

to this plane.

Using the distance formula

d = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d _{pl an e} ∣}{a ^{2} + b ^{2} + c ^{2}}

, we substitute our values:

d = \frac{∣1 ( - 1 ) + 2 ( 1 ) + 2 ( 2 ) + 7∣}{1 ^{2} + 2 ^{2} + 2 ^{2}}

Simplifying this expression yields:

d = \frac{∣ - 1 + 2 + 4 + 7∣}{1 + 4 + 4} = \frac{12}{9} = 4

Phase 3

The Geometric Bridge

We now utilize the geometric relationship between the sphere's center

O

, the circle's center

C

, and a point

A

on the circle's circumference. These points form a right-angled triangle where the hypotenuse is the sphere's radius

R

, one leg is the distance

d

, and the other leg is the circle's radius

r

By the Pythagorean theorem,

R^{2} = d^{2} + r^{2}

, which implies

r = R^{2} - d^{2}

. Substituting our known values:

r = 5^{2} - 4^{2}

r = 25 - 16 = 9 = 3

The radius of the intersection circle is 3. This journey demonstrates that even complex 3D problems can be resolved through simple, elegant geometric steps.

Common Traps & FAQs

Wait, why is the center of the sphere

(- u, - v, - w)

and not

(u, v, w)

That is a classic point of confusion! The general equation is

x^{2} + y^{2} + z^{2} + 2 ux + 2 v y + 2 w z + d_{s p h er e} = 0

. When we complete the square for the

x

terms, we get

(x^{2} + 2 ux + u^{2}) = (x + u)^{2}

. Since the standard form for a sphere is

(x - h)^{2} + (y - k)^{2} + (z - l)^{2} = R^{2}

, we identify the center as

(h, k, l)

. Comparing

(x + u)^{2}

with

(x - h)^{2}

, we see that

h = - u

. Thus, the center is indeed

(- u, - v, - w)

Can I just use the distance formula between the center and the plane to find the radius directly?

Not quite! The distance formula gives you the perpendicular distance

d

from the center of the sphere to the plane. This

d

is one of the legs of the right-angled triangle we discussed. The radius of the sphere

R

is the hypotenuse, and the radius of the intersection circle

r

is the other leg. You must use Pythagoras:

r = R^{2} - d^{2}

What if the value inside the square root for

r

turns out to be negative?

That is a great question! If

R^{2} - d^{2} < 0

, it means

d > R

. Geometrically, this implies the perpendicular distance from the center to the plane is greater than the radius of the sphere. In such a case, the plane is too far away to touch the sphere, so there is no intersection. In this problem, we found

R = 5

and

d = 4

, so

25 - 16 = 9

, which is positive, confirming the intersection exists!

Animated Solution for Mathematics - Three Dimensional Geometry: The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0$ is cut by the plane $x + 2 y + 2 z + 7 = 0$ is

Select Answer:

Visualized Solution

Visualizing the Sphere-Plane Intersection

The Geometry of the Cut

Finding the Center of the Sphere $O$

Calculating the Sphere's Radius $R$

Perpendicular Distance Formula

Calculating the Distance $d$

Finding the Circle's Radius $r$

Summary of Sphere-Plane Intersections

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The radius of the circle in which the sphere x2+y2+z2+2x−2y−4z−19=0 is cut by the plane x+2y+2z+7=0 is

Select Answer:

Visualized Solution

Visualizing the Sphere-Plane Intersection

The Geometry of the Cut

Finding the Center of the Sphere .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } O

Calculating the Sphere's Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Perpendicular Distance Formula

Calculating the Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Finding the Circle's Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Summary of Sphere-Plane Intersections

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Three Dimensional Geometry: The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0$ is cut by the plane $x + 2 y + 2 z + 7 = 0$ is

Finding the Center of the Sphere $O$

Calculating the Sphere's Radius $R$

Calculating the Distance $d$

Finding the Circle's Radius $r$

The plane $x + 2 y - z = 4$ cuts the sphere $x^{2} + y^{2} + z^{2} - x + z - 2 = 0$ in a circle of radius

The shortest distance from the plane $12 x + 4 y + 3 z = 327$ to the sphere $x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155$ is

The intersection of the spheres $x^{2} + y^{2} + z^{2} + 7 x - 2 y - z = 13$ and $x^{2} + y^{2} + z^{2} - 3 x + 3 y + 4 z = 8$ is the same as the intersection of one of the sphere and the plane

If the plane $2 a x - 3 a y + 4 a z + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres $x^{2} + y^{2} + z^{2} + 6 x - 8 y - 2 z = 13$ and $x^{2} + y^{2} + z^{2} - 10 x + 4 y - 2 z = 8$ then $a$ equals

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

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Let a plane $P$ pass through the point $(3, 7, - 7)$ and contain the line, $\frac{x - 2}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ . If distance of the plane $P$ from the origin is $d$ , then $d^{2}$ is equal to

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