Structure

5.1 Inuoduction Objectives

5.2 Equations of a Sphere 24

5.3 Tangent Lines and Planes Tangcnt Lines Tangent Planes

5.4 Intersection oC.Sphercs 3 1 Two In~ersecting Spheres Spl~eres Tlirough a Given Circle

5.5 Summary 35

5.1 INTRODUCTION

With this unit wc start our discussion of three-dimensional objects. As the unit titlcsuggests, we shall consider various aspcca ofa sphere here. A sphere is not new to you. When you wcrc a child you musl. have playcd with balls. You must also have oaten several fruits like limes. oranges and wa~rmelons. All these objccls arc spherical in shnpe. But all of thcrn are not spheres from the point of vicw of itnalylicnl geometry.

1

In this unit you will scc whal a gcomclcr calls a sphcrc. Wc shall also oblain the general cqualion ol'a sphere. Then we shall discuss linear and planar sections of a sphere. In particular, we sllall considcr thc aluations of langcnt lines and plancs 1o.a spherc. Finally, you will scc what thc intcrscction of two spheres is and how many spheres can pass through a givcn circle.

Sphcrcs are an integral part ol' the study of thc structure ol'crystals of chemical compounds. You find their properties u s 4 by architects and engineers also. Thus, an analylicul study of sphcres is not merely to satisry our mathematicnl curiosity.

A spherc is a particular case of an ellipsoid as ;';J will see whcn you study Block 3. So; if you huvc grasped the contents or this unit, i t will bc oT hclp to you while studying the next block. In othcr words, if you achievc the Following objectives, ii will beeasicr for you to understand the contents of Block 3.

w

0 bjectives

After studying this unit you should be able to

* ohmin thc cquation of a sphere if you know its centre and radius ;

* check whcthcr a given second dcgree equation in three variables represents a sphere;

0 clreck 'whelher a given line is tangent to a given sphere:

0 obtain the tangcnt plane to a givcn point on a given sphere;

0 obtain the angle of intersection of two intersecting spheres;

0 find the family of spheres passing lhrough a given circle.

The Sphere, Cone and Cylinder

Let us now see what a spherc is and how we can represent it algebraically.

5.2 EQUATIONS OF A SPHERE

In 2-space you know that the set of points that are at a fixed distance d from a fixed p in t is a circle. A sphere is a generalisation of this to 3-space (see Fig. 1).

Definition : The set of all hose points in 3-space which are at a fixed distance d from a point C (a, b, c), is a sphere with centre C and radius d.

(4 (b) I

Fig 1 : (a) A circle, (b) a spherc, with origln as centre nnd rndius d.

Spheres are, of course, not new to you. A ball and a plum are spherical in shape. However, whenevcr we mlk of a sphcre in analytical geometry, we mean the surface of a sphere. Thus, for us a hollow ball is a sphere, and a solid cricket ball is not a sphere.

Let us find the equation of a sphere with radius d and centre C (a, b, c) now. If P (x, y, z) is any point on thc sphcrc, then, by the distance formula ((I) of Unit 4). we get

which is thc requircd equalion,

For example, thc sphcre with ccnire (O,0,0) and radius 1 unit is, x2 + y? + z2 = 1.

Now, if wc cxpand (I), wc gcl the second dcgrcc equation ~ ~ + ~ ~ + ~ ~ - 2 a ~ - 2 b ~ - 2 c z + a ~ + b ~ + c ~ - d ~ = O .

! Lqoking at this you could ask iFcvcry equation of the type

where a, u, v, w, d E R, rcprcscnls a sphere. - 1

11 so happcns that if a # 0, then (2) represents a, sphere. (What happens if a = O? Unit 4 will give you the answcr.)

Lct us rewrite (2) as

u2 v2 w2 Adding + + - on either sidc of this equation, we obtain

a a a2

Comparing this with (I), we see that this is a sphere with centre The Spherc

v u2+ v2+ w2-ad, - - - - W) and radius J

a ' a 1.1

The following theorem summarises what we have said so far.

Theorem 1 : The general equation of a sphere is

la centrc is (-u, -v, -w) and radius is Ju2+ v2 + w2 - d .

Note that the general equation given above will be a real sphere iff u2 + v2+ w2 - d 2 0. Otherwise it will be an imaginary sphere, that is, a sphere with no real points on it.

So, what we have seen is that

a second degree equation in x, y and z represents a sphere iff

the coefficients of x2, y2 and z2are equal, and

ii) the equation has no terms containing xy, yz or xz.

Why don't you see if you've taken in what has been said so far?

El) Find the cenue and radius of the sphere given by x2 + y2+ z2 - 2x + 4y - 62 = 11.

E2) Does 2x2+ 1 + 2y2 + 3 + 2z2 + 5 = 0 represent a sphere?

E3) Determine the cenue and radius of the sphere x2+y2+z2 = 42.

Now, if you look at the general equation of a sphere, you will sce that i t has 4 arbitrary constants u, v, w, d. Thus, if we know 4 points lying on a sphere, then we can obtain its equation.

Lct's consider an examplc.

Example 1 : Find the equation of the sphere through the points (0, 0, O), (0, I , -I), (-1,2,0) and (l,2,3).

Solution : Suppose the equation is

x2+~'+2 + 2ux + 2vy + 2wz + d = 0.

Since the 4 given points lie on it, their coordinates must satisfy this equation. So wc get

d = O

2+2v-2w+d=O

5 - 2 ~ + 3 v + d = O

Solving this system of simultaneous linear equations (see Block 2, MTE-W), we get

Thus, the required sphere is

7(x2+y2+z2)- 1 5 ~ - 2 5 ~ - l l z = 0

'rhe Sphere, Cone end Cylinder

Note that it can happen that the system obtained by substituting the four points is inconsistent, that is, it does not have a solution. (Such a situation can occur if three of the points lie on one line.) In this case there will be no sphere passing through these pints.

You can try some exercises now.

E4) Find the centre and radius of the sphere passing throllgh (1.0.0). (0, 1,0), (O,0,1)

1 and (&, z* $)a

E5) Is there a sphere passing through (4,0, I), (10, - 4,9), (- 5,6, - 11) and (l,2,3)? If so, tinct iti equation.

A diameter of a sphere is a line segment through its centre and Now if, instead of four points on the sphere, we only know the coordinates of the two ends

with end points on the sphere, of one of its diameters we can still determine its equation. Let us see how. Let A (x,, y,, 2,) and B (x2, y2, %).be h e ends of a diameter of a sphere (see Fig. 2). Then, if P (x, y, z) is any .- -

point on the sphere, PA and PB will be perpendicular to each other. Thus, from (10) of unit 4;we see that

(x -XI) (x - - X I ) + (Y -y1) (Y -y2) + (Z-Z,) (z --z2)=0. . . . . . .(3)

This is satisfied by any point on the sphere, and hence is the equation of the sphere.

'* For example, the equation of the sphere having the points (-3,5,1) and (3,1,7) as the ends o f a d i a m e t e r i s ( x + 3 ) ( ~ - 3 ) + ( y - 5 ) ( y - 1 ) + ( ~ - 1 ) ( ~ - 7 ) = 0 ,

that is, ~ ~ + ~ ~ + z ~ = 6y + 82 - 3.

You can obtain the diameter form of a sphere's equation in the following exercise.

~ l g . 2. E6) Find the equation of the spliere described on the join of (3,4,5) and (l,2,3).

By now you must be familiar with spheres. Let us now'see when a line or a plane intersects a sphere.

5.3 TANGENT LINES AND FLANES In this section we shall first see how many common points a line and a sphere can have. Then we shall do the same for a plane and a sphcrc. .

5.3.1 Tangent Lines

Suppose you take a hollow ball and pierce it right through with a knitting needle. Then he ball and the needle will have two points in common (see Fig. 3). Do you think this is true of any line that intersects a sphere? See what the following theorcm has to say about this.

Fig. 3 : A llne intersecting a sphere Theorem 2: A line and a sphere can intersect in at most two poinw,

x- a y- b z - c proof: ~ e t x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 and - - --=-= t (say) bea a P Y

given sphere and line, respectively. Then any point on the line is of the form (at + a, pt + b, y t + c), where t E.R. If this lies on the spherq then ( a t + a ) 2 + ( p t + b ) z + ( y t + ~ ) 2 + 2 u ( a t + a ) + 2 v ( $ t + b ) + 2 w ( y t x ) c d = ~

~ ( a 2 + ~ 2 + . ; . > t 2 + 2 t ( ~ + b ~ + c y + u a + v ~ + w y ) + ( ~ + b 2 + c i + h a + 2 v b + 2wc+d)=O .. . ...(4 )

Thisis a quadratic in t. Thus, it gives two values oft. For each value oft, we will geta point 26 of interseciion. Thus, the line and sphere can intersect in at most two points.

I

Note that (4) can have real distinct roots, real coincident roots or distinct imaginary roots. ~ccordingly, the line will intersect the sphere in two points, in one point, or not at all. This leads us to the following definitions.

Definitions: If a line intersects a sphere in two distinct points, it is called a secant line to the sphcre.

I f a line intersects a spherc in one point P, it is called a tangent to the sphere at the point P; and P is called thc point of contact of the tangent.

For example, the line L in Fig. 3 is a secant line to the sphere; and the line L in Fig. 4 is a Langent to he sphere at h e point P.

Now, (4) will have coincident roots iff

(aa+ b f i + ~ ~ + u a + v P + w y ) ~ = ( a ~ + P2 + ~ ) ( a 2 + b 2 + c 2 + 2 u a + 2 v b + 2 w c + d ) ......( 5)

x - a y - b z - c Thus, (5) is the condition for - = - = t o be a tangent to a - P Y

s2+ y2+z2+2ux+2vY + 2 w z + d = 0 .

Lel us considcr an example.

Example 2: Find the interccpt made by the sphere x2+ y2+ z2 = 9 on the line x - 3 = y = z.

Solution: Ally point on the line is of the form (t + 3, t, t) where E R. This lies on the sphere if

I t - i - 3 ) 2 + t z + t 2 = 9 j 3 t 2 + 6 t = O j t = 0 , - 2 .

Thus, thc points of intersection are (3,O.O) and (1, -2, -2). Thus, the interccpt is the

dislnnce bctween the two points, which is 4- = 2&.

You can try sollie exercises now.

x i - 3 - y + 4 E7) Check if - - - = ?is a tangent to the sphere

4 3 5 x 2 + y2 + z2 + 4x + 6y + 10z = 0.

E8) 11' we extend the rule of thumb to find the ungent to a conic (see Unit 3) to a sphere, will we get the equation of a tangent line to a sphere? Why?

Lel us now discuss the intersection of a plane.and a sphere.

5.3.2 Tangent Planes Consider a sphere and a plane ha t intersects it. PYlat do you expect the intersection to be? The following result will give you the answcr.

Theorem 3: A planar section of a sphcre is a circle.

Proof: Let S be a sphere wilh radius r and centre 0 (see Fig. 5), and let h e plane Jl '

intersect it.

The Sphere

Pig. 4 : L intersects the sphere In only one palnt, P.

I

Fig. 5: A plnnar intermtion of a spherc is a circle.

The Sphere, Cone and Drop a perpendicular ON from 0 onlo n, and let ON = a. NOW ICL P be a poinl which Cylinder belongs to n as wcll as S. Thcn OP = rand 0p2 = ON' + N P ~ .

I ' Thus, NP= \/-. which is a conslanl.

Thus, the intcrseclion of S and I7 is the set of points in TI which are at a fixcd distance from

a fixed poinl N. Thus, i t is a circlc in thcplanc TI with ccntre N and radius Jn. If a = 0 in the proofabovc, Lhe planc passes through thc ccnlrc of lhc sphcrc. In lhis case the circle of intersection is of radius r and is called a great circle (sec Fig. 6) of thc spherc.

Nole that a sphere has infinilcly many grcat circles, one for cach planc lhal passes throllgh the centre of the spherc.

,,/ Wc have secn that the planar seclion o l a sphere is a cirdc': Nbw Icl us find ils equation. Lcl ~ h c equation of he sphere bc x 2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, and that of the plaric

I:ig. 6: 'She it~tcrscctin~~ r ~ f S and n inlersecling it bc Ax + By + Cz + D = 0. Tlicn he equalion or ~ h c planar seclion can bc is ;i grcat circlc V S I ~ C sphcrc S. writLen as

~ ~ + ~ ~ + ~ ~ + 2 u x + 2 v ~ + 2 w z + d = 0 = ~ x + ~ ~ + ~ z + ~ , o r I . . . . . .(6) x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, Ax + By + Cz + D = 0.

For example, lhe equalion or [he planar section of h c sphere x2 + y2 + z2 = 1 by [lie planc

1 1 2 3 z = - (see Fig. 7) is x2 + y2 + z2 - I = z - - = 0. This is lhc circlc x + y2= -, in ~ h c plane 2 2 4

1 z = - . 2

Since lhc cenlre of ~ h c given sphere is (0,0, O), wc can gct a great circle of h c sphcre by inle!.secling it will1 z = 0. Thus, onc grcat circle is x2 + y2= 1 in ~ h c planc z = 0.

Lct us considcr an cxa~nplc of he usc of Thcorcm 3.

Example 3: Find tlic ccrilrc and radius of lhc circlc 1:ig. 7: I'la~iar sect ions {,I'

s2 + y2 + 2 = I ~ ~ + ~ ~ + ~ ~ - 8 x + 4 ~ + ~ ~ . - 4 3 . 5 = 0 , x - 2 ~ + 2 z = 3 .

Solution: The cenlrc of lhc sphcre is C(4, - 2, - 3 ) and ils radius is ,

r = d l h + 4 + 1 6 + 4 5 =9. The dislancc or thc planc from lhc ccntrc of the sphcrc is

Thus, thc rrltlius ollhc circlc = dm = 4 6 .

The cenlre of thc circle is lhc fool of llic pcrpcndicular from C onio lhc planc. To find this, 1

we firs1 necd .lo find the cqualions of thc pcrpcndicular. 11s dircclion ratios arc 1, -2,2. Thus, ils equations are

Therefore, any poinl on ~lle pcrpcndicular is given by (t + 4, -2t - 2,2t - 4), whcrc t E R. This point will bc he rcquircd ccnlrc of he circlc iT it lics on thc pianc, that is, i f

13 8 Hence, thc centrc of thc circlc is

You can do the following exercise on the same lines. The Sphere

I

E9) Find the centre and radius of the circle x2+ y 2 + z 2 + 1 2 ~ - 1 2 ~ - 1 6 z + 111 = 0 = 2 x + 2 ~ + 2-17.

r

Now, if we take a = r in the p m f of Theorem 3, then what happens to the circle of intersection? It reduces to a single point, that is, a point circle. And in this case the plane only touches the sphere (see Fig. 8). .

Definition: A plane is tangent to a sphere at apoint P if it intersects the sphere in P only.

In this case we also say that the plane touches the sphere at P. P is called the point of tangency, or the point of contact, of the tangent plane.

Remark 1: If you go back to the proof of Theorem 3, you will see that thc tinc joining the point of tangency to the centre of the sphere is perpendicular to the tangent plane. We will use this fact to obtain the equation of a tangent plane.

Lci us find the equation ofjhe tangent plane to Llle sphcre x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 at the point P (a, b, c).

As P lies on the sphere,

Also, the cdntre of the sphere is C (-u, -v, -w). Thus, the direction ratios of CP are a + u, b + v, c + w (see Equation (8) of Unit 4). .

Now, the tangent plane passes through P (a, b, c). Thus, ih quation will be

f (x - a) + g (y - b) + h (z - c) = 0, for some f, g, h E R. . .. .. .(8)

Now CP is perpendicular to (a), and hence, is parallel to the normal to (8). Further, f, g, h :\re direction ratios of the normal to the plane. Therefore, a + u, b + v, c + w and f, g, hare proportional.

f g h . . ' - = - = = t, say. a + u b + v c + w

Thcn (8) gives us

(x -il)(a + u)+ (y-b) (b+ v)+ (z-c) (c + w) =0.

~ x a + y b + z c + u x + v y + w x = a 2 + b 2 + c 2 + u a + v b + w c . . . . .. .(9)

llsing (7) and (9), we get

x a - + y b + z c + u x + v y + w z = - u n - v b - w c - d .

Thus, '

the equation of the tangcnt plane at the point (a, b, c) lo the sphere* ~ ~ + ~ ~ + ~ ~ + 2 u x + 2 ~ ~ + 2 w z + d = 0 i s x a + y b + z c + u ( x + a ) + v ( y + b )+w(z+c)+d=O. I 1

Is this the equation you got whilc doing Ell? From the equation you may have realised that there is a similar thumb rule for the langcnt plane (and not tangent line!) to a sphere.

Fig. 8: The plane n Is tangent to the sphcrc S at the point N.

Rule of Thumb: To obtain the equation of the tangent plane to a sphere at the point (a, b, c), simply substitutc ax for x2, by for y2, cz for z2; and in the linear terms substitulc x + a . - z + c +

for y, - for z in the ec$pion of the sphere. 2

for x, - 2 2

'The Sphere, Cone and Cylinder

For example, the equation of he tangent plane to x2 + y2 + z2 = a2 at (a, P, y) is xa + yp + zy=a2.

Why don't you try an exercise now?

- -- - - - - - - - -

El 0) Find the equations of the tangent planes a ) ~ o x ~ + ~ ~ + z ~ + 2 ~ = 2 9 a t ( 2 , 3 , - 4 ) . b) tox2+ y 2 + z 2 - 4 y - 6 z + 4 = ~ a t ( 2 , 3 , 1).

So, what we havc secn so far is that if a plane is at a distance d from the cenae of a sphere with radius r, thcn

i) if r < (1, the plane and spherc do not intersect;

ii) if r = d, the plane is tangent to the sphere; and

i i i ) if r > d, the plane intersects h e sphere in a circle of r a d i u s ' d n .

Now, if yo1 are given tlie Quations of a sphere and a plane, can you tell if the plane is tangent to .the spllere'? An obvious way would be to check what the distance of the centre of the sphere from the plane is. Let us use this method to derive the condition for the plane Ax + By + Cz + D = 0 to be a tangent plane to the sphere x2 + y2+ z2+ 2ux + 2vy + 2wz+d=O.

Now, thc radius of the sphere is du2 + v2+ w2 - d .

Thc length of the perpentlicular to the plane from the centre (-u, -v, -w) of the sphere is

This dislancc must equal lhc sphere's radius since Lhe plane is tangent to the sphere.

:. (Au + Bv + Cw + D ) ~ = ( A ~ + Fi2+ c2) (u2 + v2 + w2 - d), ......( 10)

which is the requircd condilion.

Let us consider an ekample.

Example 4: Show that 2x - y - 22 = 16 touches the sphere x Z + y2 + z2 - 4x + 2y + 22 - 3 = 0, and find the point of contact.

Solution: The ccntre ol'ihe sphere is (2: -1, -1) and its radius is d m = 3.

Thc lenilh of the perpendicular from the ccnue to the plane 2x - y - 22 - 16 = 0 is

/2.2+1+2-161 9 = - = 3, which is thc same as the radius of the sphere.

d m 3

So the plane iouches thc sphere.

Let (x,, y,,, zl) be the poinl of contncl. Then the equation of [he tangent plane is

X X , + yy, + ZLl - 2(x + x,) + (y + y,) + (2 + 2,) - 3 = 0

~ ~ x , - 2 ) x $ ( y , + 1 ) y . C . . ( z , t . 1 ) z - 2 x , + y 1 + z , - 3 = 0 .

But this Should bc the same 3s thc given plane 2x - y - 22 - 16 = 0.

So the coefficients of x, y, z and the constant term in both these equations must be proprtional.

'. xl = -2yl, zl = 1 + 2yl, and then

) :. x, = 4 and zl = -3. ri

Thus, the point of contact is (4, -2, -3).

Using the same method, if we are given a point and a plane, we can find the sphere with the poidt as the centre and the plane as a tangent plane. In the example below we illustrate this.

i Example 5: Find the sphere w h centre (-1,2,3), and which touches the plane 2x - y + 22 = 6.

Solution: The distance of the plane from the point is

This should be the radius of the sphere.

So, sincc the centre of the sphere is (-1,2,3), its equation will be

Why don't you do some exercises now?

Ell) Show that the plane x + y + z = & touches the sphere x2 + y2 + z2 = 1. Find the

point olcontact.

E12) Show that the equation of thc sphcre which lies in the ocunt OXYZ and touches the coordinate planes is x2 + y2 + z2 - 2k (x + y + z) + 2k2 = 0, for some k E R.

E13) Find the equation of the spherc with centre ( I , 0, O), and which touches the plane 2 x + y + z - 3 = 0 .

In this scction we have seen what sets can be got by intersecting a line or a plane wi~h a sphere. Now let us discuss what form the inkrsection of two or more spheres can take.

k 5.4 INTERSECTION OF SPHERES 1

In this section you will lirst see that the result or intersecting two spheres is the same as that . obtained by intersecting a sphcre and a plane, that is, a circle. And thcn you will see how to obtain infinitely many spheres whose intersection is a given circle.

5.4.1 Two Intersecting Spheres

Let us consider two spheres given by

SI = x2 + y2 + z2 + 2u1x + 2vly + 2wlz + dl = 0, and

' Then each point that satisfies S1 = 0 as well as S2 = 0 will also satisly the equation S1 - Sz = 0, that is,

The Sphere

i

Thus, thc spheres S, = 0 and S2 = 0 intersect at 90' iff (13) is satisfied. The Sphere

Let us consider an example. i

Example 6: Find the angle between x2 + y2 + z2 = 4 and x2 + y2 + z2 - .2x = 0.

Solution : Here u l =O,vl = 0 , wl= O,dl = -4,u2=-1,v2=0, w2 =0, d2=0.

Thus, the centres of the two spheres are (0,0,0) and (1,0, O), their radii are 2 and 1, rcspectivcly, and the distance between their centres is 1.

Therefore, by (12), the angle between the two spheres is

Fig. 11

You can see these spheres in Fig. 11. They intersect in only onc point P, and the x-axis is the normal from the centres of the spheres to both tangent planes.

You can try some exercises on intersccting spheres now. 1

E13) Find the angle of intersection of the spheres x2 + y2 + z2 - 2x + 2y - 42 + 2 = 0 and x2.+ y2 + z2 = 4.

E15) Find the equation of the sphere touching he planc 3x + 2y - z + 2 = 0 at P (1, -2.1) and cutting the sphcre x2 + y2 + z2 -4x + 6y + 4 = 0 orthogonally,

€16) a) Two spheres of radius rl and r2 and with centres C1 and C2, respectively, will touch each other iff rl + r2= C1C2. Truc or false? Why?

b) Undcr what conditions on rl, r2 and ClC2 will thc spheres not intersect?

E17) ~ h o w t h a t t h c s ~ h c r c s x ~ + ~ ~ + z ~ - 2 ~ - 4 ~ - 4 ~ = 0 a n d x ~ + y ~ + z ~ + 10x+22+ 10 = 0 touch each other. What is the point of contact?

So far you havc seen that two sphercs intersect in a circlc. Now let us see whether, given a circle, wc can find LWO or more sphercs passing through it.

5.4.2 Spheres Through a Given Circle / I Supposc we are given a circle. Can we find two distinct sphercs whose intersection the circle is'? In fact, we can construct many spheres passing through a given circle (scc Fig. 12).

Fig, 12: Part nf a family of sphercs through n clrcle.

In Fig. 12, the circle is a great circle of the sphen: S1, but not of Sz, SS, etc. Let us see what the method of construction of this kind of family is.

You know that a circle is the intersection of a sphere and a plane. So its equation is of the form S = 0, I7 = 0, where

S = x 2 + y2 + ~ ~ + 2 u i ( + 2 ~ ~ + 2 w z + d , a n d I I = ~ x + ~ ~ + ~ z + D.

'l'he Sphere, Cone and Cylinder

Any sphere through this circle will be given by

S+kn=O, ......( 14)

where k is an arbiuary constanl. Do you agree? Now, if you apply Theorem 1, you can see that (14) represents a sphere.

Furlher, every point that lies on the circle must satisfy (14). Thus, (14) represents a sphere lhrough the given circle.

So, for each value of k E R in (14) we get a distinct sphere passing through the given circle. Thus, we have infinitely many spheres that intersect in h e given circle.

Now, a circle can also be represented as the intersection of two spheres S1= 0 and S2 = 0. In his case what will the equation of any sphere containing it be? It will be S1 + kS2 = 0, where k E R. Thus, the infinite set [S1 + kS2 = 0 I k E R j gives us the family of spheres passing through the given circle.

Let us consider some examples of the use of (14).

Example 7: Show that the circles x2 + y2 + z2 - y + 22 = 0, x - y + z - 2 = 0 and

i 2 + p 2 + x 2 + ~ -3y+z-5=0 ,2x-y+4z- 1 =Olieonasphere,andfind itsequation.

Solution: Tho equation of any sphere through the first circle is

~ ~ + ~ ~ + ~ ~ - ~ + 2 ~ + k ( x - ~ + z - 2 ) = 0 , t h a t i s ,

x2+y2+z2+kx-(k+ l )y+(k+2)z -2k=0 , ...,.. (15)

for some k E R.

Similarly, be equation of any sphere lhrough he srcond circle is

x2 + y2 + z2 + (2kl + 1) x - (k, + 3) y + (4k1 + 1) z - (kl + 5) = 0, . .... .(16)

Tor some k, E R.

To get a cornmon sphere containing bolh circles, we must see if (I 5) and (16) coincide for somc k and k1 in R. Comparing lhe coefficients of x, y and z, and lhe constant terms in (15) and (I(,), wc get

Thcse cqualions arc salisl'icd lor k = 3 and kl = 1.

Thus, thcrc is a sphere passing through both the circles and ils equation is

Example 8: Find thc equalion oC the sphere through the circle x2 + 9 + z2 = 9,2x + 3y + 41. = 5 I

2nd thc origin.

Solution: LQ the equation of the sphere be x2 + y2 + z2 - 9 + k (2x + 3y + 42 - 5) = 0, where k € R.

9 Since, it passcs through (0, 0, 0), wc get -9 - 5k = 0, thal is, k = - - . 5

Thus, the rcquired equation is

5(x2 + y2 +z2) = 9 ( 2 ~ + 3y 4- 42).

Example 9: Find the path traced by the centre of a sphere which touches the lines y = x, Z = landy=-x, z=-1.

Solution: Let x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 be the equation of a sphere that I

touches ~ h c 2wo lines. Since y = x, z = 1 touches it, the intersection of the line and the sphere must bc only one point. Any point on the line is (t, t, I), where t E R. It lies on the sphere if

t2+t2+ 1+2ut+2vt+2w+d=0.

I 1 I

11

This equation has equal roots if

(u+v)'=2(1+2w+d). '

Similarly, since y = -x, z = -1 touches the sphere, we get

(u - v12 = 2 (1 - 2w + d).

Subtracting these two conditions, we get

4uv = 4w (1 + l), that is, uv = 2w.

Thus, the centre of the sphere, (-u, -v, -w), satisfies the equation xy + 22 = 0.

This is true for any sphere satisfying the given conditions. Thus, the required path is xy-k 22 = 0.

Now, why don't you check if you've understood what we have done in this section so far?

E18) Prove that the circles

x2+y2+z2-2x+3y + 4 z - 5 = 0 , 5 y + 6 ~ + 1 = 0 a n d

~ ~ + ~ ~ + z ~ - 3 ~ - 4 ~ + 5 z - 6 = 0 , x + 2 ~ - 7 z = 0

lie on the same sphere. Find its equation also.

€19) Find the equations of the spheres that pass through x2 + y2 + z2 = 5 . 2 ~ + y + 32 = 3 and touch the plane 3x + 4y = 15.

E20) Find the equation of the sphere for which the circle 2x - 3y + 42 = 8, x2 + y2 + 2 + 7y - 22 + 2 = 0 is a great circle.

We will stop our discussion on spheres for now, though we shall refer to them off and on in the ncxt block. Let us now do a quick review of what we have covered in this unit.

5.5 SUMMARY

The Sphere

In his unit we have covercd the following points.

1) The second degree equation . x Z + y 2 + z 2 + 2 ~ x + 2 v y + 2 w ~ + d = o

represents a sphere with centre (-u. -v. -w) and radius du2 + v2 + w2 - d . Conversely, the equation of any sphere is of his form.

2) A line intersects a sphere in at most two points. It is a tangent to the sphere it' it intersects the sphere in only one point.

3) A plane intersects a spherc in a circle. When this circlc reduccs to a point circle P, then the plane is tangent to the sphere at P.

4) The equation of the tangent plane to the sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 at h e point (x,, yl, zl) is xx, + yy, + z2, + u (x + x,) + v (y + y,) i- w (z + z,) + d = 0. This is perpendicular to the line joining (x,, y,, z,) to the centre or the sphere.

5 ) Two spheres intersect in a circle,

6) The angle of intersection of the two intersecting spheres x2+ y2 + z2+ 2ulx + 2vly + "

2w,z + d, = 0 and xZ + y2 + z2 + 2u2x + 2vzy + 2w2z + dz = 0 is cos-' rf -t r$ - d2

where r1 and r, are their radii and d is Ihc distance between their centres.

In particular, the two spheres are srlhsgonal if 2u1u2 + 2v1v2 C 2wIw2 = dl + d2

' l ' l~e Sphere, Cone ilnd Cylinder

7) Thcrc arc infinitely many sphcres that pass through a givcn circle.

You may now like 6 go back lo Sec. 5.1 and go through the list of unit objectives to see if . you have achievcd them. 1r you wan1 to see what our solutions Lo the exercises in Ihe unit

arc, wc have givcn thcnl in ~ h c li)llowing seclion.

Its centre is (-($1, - (4). -($)')=(I. - 2, 3).

11s radius is ,/(-1)2 + + (-312 - (-1 1) ='5.

9 E2) Wc can rcwrite thc equation as x2 + y2 + z2 + - = 0.

2 This represents an imaginary sphere wi~h centre at the origin.

E3) I t s centre is (0, 0,2) and radius is 4.

4 ) Lcl the spliere bc x 2 + y2 t z2 + 2ux + 2vy + 2wz + d = 0. Then, since thc given poinls lie on il, wc get 1 + 2 u + d = O 1 + 2 v + d = O 1 + 2 w + d = O

On solving thcsc equations, we find that u = v = w = 0, d = -1. Thus, llic ccntrc ol'llie sphere is (O,0,0) and radius is 1.

E5) Suppose such a spliere exists. Let ils cquation be ~ ~ + ~ ~ + z ~ + 2 u x + 2 v ~ + 2 w z + d = 0 . Sincc the givcn points lie on it, we get the linear system 17+8u+2w+t i=O 197+2Ou-8v+ 18w+d=O 192 - IOU + 1 2 ~ - 2 2 ~ + d = 0 14+2u+4v+Gw+d=O. You can check ha this systcm is inconsistent. Thus, the poink do not lie on a sphere.

E ) 'The required equation is (x-3)(x - l )+ (y -4 )o ' - 2 )+ (z - -5 ) ( z -3 )=0 . e x 2 t y 2 + z Z - 4 x - 6 ~ - 8 ~ + 2 6 = 0 .

€7) Any point on the line is (4t - 3,3t - 4,5t), where t E R. This will lie on the sphere if (3t -312 + (3t -412 + 25t2 + 4 (4t - 3) + 6(3t -4) + 10 (5t) = 0. w Sol2+ 36t- 11 = 0

Since these are real distinct roots, thc line will intersect the sphere in two distinct poinb. Hence, it will not be a langcnt to the sphere.

E8) If we extend the rule or lliumb to obhin the tangcnt at a point P (x,, yl, zl) on the s p h e r e x 2 + y 2 + ~ + 2 u x + 2 v y + 2 w z + d = ~ , w e g e t . xxl + yyl+ zz, + u (X + xl) + v(y f yl) + w(z + zl) + d = 0. This is a linear equation, and hence'rcpresents a plane, and not a line. Thus, it cannot represent the tangent line,

~ 9 ) The centre of h e sphere is C (- 6,6,8). Its radius is r = 5. The dislance of C from the plane is d = 3.

Thus, the radius of the circle = = 4.

x + 6 y - 6 The equations of the perpendicular from C onlo the plane are - = - = , - 8. 2 2

Thus, the centre of the circle is (- 4,8,9).

El 1) The radius of the sphere = 1: The distilnce of the plane from he centre (0,0, 0) of Lhc sphere = 1. 'Thus, the plane is tangent to h e sphere. If the point of contact is (a, b! c), then the equation of the plane is ax + by + cz - 1 = O,as well as x + y + z - & = 0.

a b c 1 . -=-=-=- . . 1 1 1 &.

Thus, lhe point of contact is (& $1 $1. E12) Let h e cquation bc

X ~ + ~ ~ + ~ ~ + ~ U X + ~ V ~ + ~ W Z + ~ = O .

Since the plane x = 0 is a mngent lo it, the distance of (-u, -11, -w) from x = 0 musl

be d u 2 + v2 + w2 - d = r, SiLy.

(Nole thal I - u I = - u, since lhe centre lies in the oclanl in which the x, y and z coordinalcs are all posi~ivc.) Similarly, -v = -w = r, Thcn u2+v2+w2-d '=r2+d=2r2. Thus, Lhe equation of the sphere is x ~ + ~ ~ + z ~ - 2 r ( ~ + ~ + z ) + 2 ? = 0 .

12-31 1 E13) Its radius should be - = - & 47,

Thus, its equation is

El 4) Thcir centres are C, (1, -1,2) and C2 (0,0, O), rcspectivcly. Boll) Lhcir radii arc 2, ant1 c ~ c ~ ~ = 6. Tltus, thc angle of inlcrscction is

.El 5) Lcll thc sphcre hc given by x2,+ Y2 + x2 + 2ux + 2vy + 2wz + d = 0. Then the plane 3x + 2y + z + 2 = 0 is the same as x - 2 y + x + u ( x + l ) + v ( y - 2 ) + w ( x + 2 ) + d = O e x ( 1 + L I ) + ~ ( v - ~ ) + z ( w + ~ ) + u - ~ v + w + ~ = O

The Sphere