Vectors and Geometry Part-13

VECTORS AND GEOMETRY PART-13

SPHERE

PREPARED:RASHMI

PRESENTEDAND CONTENT EDITED:NANDAKUMAR.M`cz

OBJECTIVES

1.Studies the concept of sphere

2.Familiarise different equations of sphere in the different situations

3.Studies the intersection of two spheres

Introduction

In this session we begin with the definition of a sphere ,discuss equations of a sphere in various situations .Here we introduce the concepts of Plane section of a sphere,intersection of two spheres,greatcircle.Alongwith these we discuss problems involving the concepts.

definition : A Sphere is the locus of a point in space which moves so that the distance from a fixed point always remains a constant. The fixed point is called the centre of the sphere and the constant distance is called the radius of the sphere

1.Equation of a sphere with given centre and radius

Let be a fixed point and be any point on the surface of a sphere of radius , where is a constant (see figure 1).Then, we have the equation of the required sphere as

. . . (1)

Fig:1

Equation (1) can also be written as x2 + y2 + z2 - 2ax - 2by - 2cz + (a2 + b2 + c2 - r2) with centre (a,b,c) and radius r

Remark

1.Equation of a sphere is a second degree equation in x,y and z

2.Coefficients of x2 ,y2and z2 are equal

3.The product term involving xy , yz and xz are absent

Corollary :If the centre of the sphere is the origin , the corresponding equation of the sphere is . . . (2)

General equation of a sphere

Consider the equation

. . . (3)

Then completing the squares, we can write (3) in the form

The above can be re-written in the form

... (4)

Comparing Equation (1) and (4), we obtain

2. Equation of a sphere with join of two given points as diameter

Let be the diameter of a sphere with and as the extremities of the diameter. Also let be any point on the surface of the sphere(see fig 2).

Now, the direction ratios of are .

Similarly, the direction ratios of are .

Then, since, we have

Sphere through four given points

To find the equation of the sphere passing through the points and .

Let the sphere be… (5)

If it passes through, , we have

… (6)

… (7)

… (8)

… (9)

Eliminating from (5), (6), (7), (8) and (9), we get determinant of

which is the required equation of the sphere.

Example 1. Find the equation of the sphere with centre at and radius units.

Solution The required equation is

i.e.

Example 2: Find the coordinates of the centre and the radius of the sphere

Solution:The given equation is

Dividing throughout by 2,

i.e.,

Hence centre of the sphere is and its radius is .

Example 3: Find the equation of the sphere described on the line joining the points and as diameter.

Solution:The required solution is i.e.,

Example4 :Find the equation of the sphere passing through the origin and makes intercepts on the axes.

Solution:

Let . . . (10)

be the equation of the required sphere. Since the sphere passes through the origin and makes intercepts on the axes, the points lie on the sphere (10).

Hence

and solving above equations we have , ,

Substituting in (10), the required sphere is

Example 5: Find the equation to the sphere through the point, , , .

Solution:Let the equation of the sphere be… (11)

It passes through … (12)

Also it passes through other three given points.

or… (13)

or… (14)

and

or… (15)

Adding (14) and (15),

… (16)

Solving (13) and (16)

And

Then from (14),

Putting these in (11), the equation of the sphere becomes

3. Plane section of a sphere

Here we show that every plane section of a sphere is a circle and to find the radius and centre of the circle so obtained.

Let O be the centre of the sphere and ABC a plane section of the sphere.

Take a point P on the section and drop ON perpendicular on the plane. Let N be the foot of perpendicular see fig (3).

Obviously, ON is perpendicular to NP and OP is radius of the sphere.

Hence constant, since O and N are fixed.

Hence locus of P is a circle, whose centre is N and radius PN.

Clearly, centreN is the point of intersection of the plane ABC with a normal to it through centreO of the sphere.

Intersection of two spheres

To show that the centre of intersection of two spheres is a circle.

Let the two spheres be given by

… (17)

… (18)

Coordinates of points on curve of intersection of (17) and (18) satisfy (17) and (18) both, and also the equation

… (19)

which is the equation of first degree and hence represents a plane.

Thus the curve of intersection of two spheres is the same as that of a plane (19) with any of the spheres. Hence by the curve of intersection of two spheres is a circle.

Example 6: Obtain the radius and centre of the circle

Solution

Centre of the given sphere is and radius is . Also the length of the perpendicular from the centre to the given plane.

… (20)

Thus, and

radius of the circle

Further, the line through centreO and perpendicular to the given plane (20), is

… (21)

Centre of the circle will be the point where the line (2) meets the plane (20).

Let the line (21) meets the plane (20) at the point . Substituting this point in (20), we get

Hence the point N will be . This is the required centre of the circle.

Example 7Find the equations to the spheres through the circle , and (a) the origin, (b) the point .

Solution

Any sphere through the given circle is

(a) If it passes through the origin, then

And then in this case above equation becomes

or.

(b) If the spherepasses through , then

Hence the required equation of sphere becomes

or.

Great circle

The section of a sphere by a plane passing through its centre is known as a great circle .The centre and radius of a sphere are the centre and the radius of the great circle.

Example 8: Prove that the plane cuts the sphere in a circle of radius unity and find the equation of sphere which has this circle for one of its great circle.

Solution

The centre of given sphere is and its radius is

Also, length of perpendicular from to the plane

Radius of circle

Now the equation of a sphere through given circle is

Its centre is

If the circle is a great circle of the sphere above, then its centre should lie on the plane of the circle.

the equation of required sphere is

Example 9:Show that the sphere

cuts

in a great circle if

or,

where is the radius of the second sphere.

Solution:

The plane of the circle, i.e., the plane in which the circle of intersection lies, is ,

The circle of intersection will be the great circle of the sphere only when the above plane passes through the centre of the sphere , i.e., passes through the point . Hence

or… (22)

Now, we have

from (22) we get

Hence the result

Summary

Now let us summarise what we have discussed .in this session first we introduced equation of a sphere in different ways with numerous examples . Intesection of two spheres,plane section of a sphere are also discussed .

Assignment

Find the equations of the circle which lies on the sphere and has its centre at (3, 3, 3)
Find the equation of the sphere through the circle

and the point (2, 1, 1)

Find the equation of the sphere whose centre is (2,-3,4) and radius 5 units .

FAQ

1.Is there any formula for finding the coordinates of the centre and radius of a sphere

Ans:consider the general equation of the sphere

this equation can be written as

hence the centre is (-u,-v,-w) and radius is

QUIZ

1.The radius of a point circle

(a) 1(b)0(c)2(d) infinite

2.Centre of the sphere

(a)2,-1,1(b)-4,2,-2(c)-2,1,-1(d)4,-2,2

Answers :1 (b) 0

2(a) 2,-1,1

GLOSSARY

SPHERE:A Sphere is the locus of a point in space which moves so that the distance from a fixed point always remains a constant.

GENERAL EQUATION OF A SPHERE:,with centre and radius (-g,-f,-h) and respectively.