VECTORS AND GEOMETRY PART -15

Properties of sphere

PREPARED: RASHMI

PRESENTED AND CONTENT EDITED :NANDAKUMAR

OBJECTIVES

1.Studies properties of sphere

2.Introduces the Radical plane

3.Familiarise the conditions of tangency

INTRODUCTION

In this session we discuss equation of a tangent plane at a point on the sphere,condition of tangency,some of the properties of sphere, and the idea of radical plane.

1. Equation of a tangent plane at a point

To find the equation of the tangent plane at a point p(x1, y1, z1) on the sphere ……….(1)

The point p(x1, y1, z1) lies on

(1) hence

……………(2)

Any line through the point (x1, y1, z1) is

…………(3)

Where l ,m,n are actual direction cosines .if (3) meets

(1) in points then the distance of points of intersection from p are given by

One value of r is zero which shows that p is a point on the

sphere .now the line (3)will be tangent line if the other value of r also coincide with p(x1, y1, z1) .the other value of r will be zero if ……………. (4)

Thus (3) will be a tangent line to (1) if it satisfies condition (4).to obtain the locus of line (3)for different values of l,m,n we will eliminate l,m,n from (3) and (4) .obviously this locus will be the tangent plane to the sphere at (x1,y1,z1) .eliminating

l,m,n from (4) with the help of (3) we get

(x-x1)(x1+u)+(y-y1)(y1+v)+(z-z1)(z1+w)=0 or xx1+yy1+zz1+ux+vy+wz=x12+y12+z12+ux1+vy1+wz1

adding ux1+vy1+wz1+d to

both sides and using (2)we get

xx1+yy1+zz1+u(x+x1) +v(y+y1)+w(z+z1)+d=0.

this is the equation of tangent plane at (x1,y1,z1)

Remark:To write equation of a tangent plane ,replace x2 to xx1

y2 to yy1,z2 tozz1,2x to

x+x1,2y to y+y1 and 2z to z+z1 in the equation of the sphere

Theorem 1:To prove that the tangent plane at any point of a sphere is perpendicular to the radius through that point

solution: Let the equation to the sphere be

……..(5)

Equation of the tangent plane at a point (x1,y1,z1) is

xx1+yy1+zz1+u(x+x1)+v(y+y1)+w(z+z1)+d=0,hence direction ratios of a normal to the tangent

plane (5) are x1+u,y1+v,z1+w .again coordinates of the centre of the sphere are (-u,-v,-w).hence the direction ratios of the line joining the centre (-u,-v,-w)to the point (x1,y1,z1) are x1-(-u),y1-(-v),z1-(-w) ie

x1+u,y1+v,z1+w . these are the same as that of the normal to the tangent planes . hence the

tangent plane is perpendicular to the radius through the point of contact.

Theorem2: To show that the tangent line to a sphere is

perpendicular to the radius through the point of contact.

Solution: let the tangent line at (x1,y1,z1) be

This will be a tangent line to the sphere

if =0

this relation shows that line whose direction cosines are l,m,n is at right angles to a line whose direction ratios are x1+u,y1+v,z1+w ie the radius through the point of contact.

Condition of tangency

To find the condition that the

plane lx+my+nz=p is a tangent plane to the sphere

Centre of the given sphere is

(-u,-v,-w) and its radius is

If plane lx+my+nz=p touches the sphere , then perpendicular distance of the

plane from the centre of the sphere must be equal to its radius ie

Or (Lu+mv+nw+p)2 =(L2+m2+n2) (u2+v2+w2-d) this is the required condition

Example1:Find the eqation of a sphere which touches the sphere

at (1,2,-2) and passes through the point (1,-1,0)

Solution: Equation of tangent plane to the given sphere at (1,2,-2) is

X(1)+y(2)+z(-2)+(x+1)-3(y+2)+1=0 or 2x-y-2z-4=0

Let the required sphere be

As it passes through the origin so or .hence the equation of the required sphere be

or

**2.properties of a sphere**

Theorem3:The condition that the plane a x + b y + c z + d = 0 may touch the sphere is .

Proof : If the given plane is a tangent plane to the sphere, then the length of the perpendicular *p drawn from the centreC(-u, -v, -w) is equal to the radius r *of the given sphere. That is .

But

Hence the condition for tangency is

.

Remarks

- The radius of the circle of intersection of the sphere by the plane is given by

- The length of the tangent line drawn from to the sphere S = is given by

which can also be written as

Definition :If C is the centre of a sphere and r, its radius and P, any point in space, then is called the power of the point P with respect to the sphere . Hence the power of with respect to the sphere

is

which is positive or negative according as P lies outside or inside the sphere.

Theorem 4: The vector equation of a sphere

at the point A whose position

vector is

………(6)

Proof : Let C be the centre and its position vector be c. Let P be any point on the plane through and perpendicular to CA . Let its position vector be r. Now CA is perpendicular

to AP. So the equation of this plane is

or (7)

The condition that A is a point on the sphere is obtained by setting

in equation (6) and is

(8)

So addition of (7) and (8) gives the equation of the tangent plane as

or

i.e.,

Corollary 1 :Condition for the plane , where n is the unit vector , to touch the sphere is

or

or

Example2: A sphere of constant radius 2k passes through the origin and meets the axes in A,B,C. Find the locus of the centroid of the tetrahedron OABC

Solution: Let Co-ordinates of the points be and respectively.

Then equation of the sphere

OABC is

Radius of this sphere is given equal to 2k.

Let be the coordinates of the centroid of the tetrahedron OABC; then

, so that

Eliminating from the above equation , the required locus is

or.

Example 2 :A sphere whose centre lies in the positive octant passes through the origin and cuts the planes , in circles of radii

respectively show that its equation is

Solution

Since the sphere passes through the origin, let its equation be

Plane cuts it in

The radius of this circle is .

Similarly,

And

These gives

Substituting the values of and w in , we get the required equation.

Radical plane

Suppose that the equations of the two given spheres are

If is a point which moves such that the tangents from it to the spheres are of equal length, then the equation of the locus of P is obtained from

as or

This represents a plane to the line of centres. This plane is called radical plane of the spheres. The radical plane also can be defined as the locus of a point which moves such that its power with respect to the spheres are equal.

Summary

Now let us summaries what we have discussed ,here we discussed equation of a tangent plane at a point on the sphere with supporting theorems .condition of tangency is also discussed, at last we moved to some of the properties of sphere together with radical planes.

Before moving to the next session let us try these questions

Hope u enjoyed the session see u next time till then good bye

ASSIGNMENT

- Obtain the equation of the tangent plane at the origin to the sphere
- Prove that the sum of the squares of the intercepts made by a given sphere on any three mutually perpendicular lines through a fixed point is constant
- Find the locus of the centre of the sphere of constant radius which pass through a given point and touch a given line .

FAQ

1.Find the equation of spheres which pass through the circle

And touch the plane

Solution;

Equation of a sphere through the given circle is

This will touch the plane 3x+4z=16 if distance of plane from centre

=radius

Ie

This will give two values of

Corresponding to each value there being a sphere

2. How two spheres touch internally?

Solution ;Two spheres touch internally if the difference of their radii is equal to the distance between their centers

GLOSSARY

POWER OF THE POINT:If C is the centre of a sphere and r, its radius and P, any point in space, then is called the power of the point P with respect to the sphere

QUIZ

1.The centre of the sphere

(a) (-8,6,-4)(b)(-4,3,-2) (c) (8,-6,4) (d)(16,12,8)

2.The radius of the sphere

(a)45(b)36(c)25 (d)100

Answers

1 (b) (-4,3,-2)

2(b)36

Reference

1S.L.Loney The Elements of Coordinate Geometry ,Macmillian and company, London

2Gorakh Prasad and H.C.Gupta Text book of coordinate geometry, Pothisala pvt ltd Allahabad

3P.K.Mittal Analytic Geometry Vrinda Publication pvt Ltd,Delhi.