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.