If the center of the sphere is the origin the the equation is. x y z 2ux 2vy 2wz d 0 -(2)

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Felix Patterson
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1 Sphere Definition: A sphere is the locus of a point which remains at a constant distance from a fixed point. The fixed point is called the centre and the constant distance is the radius of the sphere. Let C(x 0, y 0,z 0) be the centre and r be the radius of a sphere S. Consider a point P(x, y,z) on the sphere. Then the equation of the sphere is given by (x x ) (y y ) (z z ) r - (1) If the center of the sphere is the origin the the equation is x y z r. General equation: Expanding (1), we get x y z 2x x 2y y 2z z x y z r This equation is of the form x y z 2ux 2vy 2wz d 0 -(2) where u x, v y, w z, d x y z r Since C(x 0, y 0,z 0) is the centre and r is the radius of the sphere (2) represents the equation of the sphere whose center is ( u, v, w) and radius is u v w d. Example 1. Find the equation of the passes through the point (1,-2,0). Solution. Since (3,-1,4) is the centre, equation is The sphere passes through the point (1,-2,0), we have (1 3) ( 2 1) (0 4) r r 21 sphere whose centre is (3,-1,4) and which (x 3) (y 1) (z 4) r The required equation is (x 3) (y 1) (z 4) Obtain the equation of the sphere which passes through the points (1,0,0), (0,1,0), (0,0,1) and which has its centre on the plane x y z 6. Solution. Let x y z 2ux 2vy 2wz d 0 be the required equation. Since it passes through the given points, we have

2 1 2u d 0 1 2v d 0 1 2w d 0 1 u v w (d 1). 2 Its centre lies on the plane x y z 6. u v w 6 d=3. Required equation is x y z 4x 4y 4z a) Find the equation of the sphere which has (x 1, y 1,z 1) and (x 2, y 2,z 2) as the extremities of a diameter. b) Find the equation of the sphere having the points (2,1,-3) and (1,-2,4) as the ends of a diameter Find its centre and radius. Solution. Consider a point P(x,y,z) on the sphere S having the points A (x 1, y 1,z 1 ) and B (x 2, y 2,z 2 ) as the extrimities of a diameter. Then AP and BP are at right angles. The direction ratios of AP are (x x 1, y y 1,z z 1) and those of BP are (x x 2, y y 2,z z 2). D A B C (x x 1)(x x 2) (y y 1)(y y 2) (z z 1)(z z 2) 0 is the required equation of the sphere. b) Given A=(2,1,-3), B= (1,-2,4) Equation is (x 2)(x 1) (y 1)(y 2) (z 3)(z 4) 0. x 2 y 2 z 2 3x y z 12 0

3 3 1 1 Centre= u, v, w,,. And radius= u v w d = Intersection of a plane and a sphere: Figure Section of a sphere by a plane is a base circle and the section of a sphere by a plane through its centre is called a great circle. The equations And x y z 2ux 2vy 2wz d 0 Ax By Cz D 0 taken together represents a circle. Having center M and the radius 2 2 MA r p. Sphere through a circle of intersection The equation of a sphere that passes through the circle of intersection of the sphere and the plane is given by S+kT=0 where S=0 is the equation of the sphere and T=0 is the equation of the plane. Example 1. Find the centre, the radius and the area of the circle x y z 2y 4z 11, x 2y 2z 15. Solution. Centre of the sphere, C (0,1, 2)

4 And radius, r Let M(x 1, y 1,z 1) be the centre of the circle. Then the direction ratios of CM are (x 1, y1 1,z 1 2) Since CM is perpendicular to the plane x1 y1 1 z1 2 t M (x 1, y 1,z 1) (t,2t 1,2t 2) Since M lies on the given plane, we have x1 2y1 2z1 15 t 4t 2 4t 4 15 t=1 M (1,3,4) and p Radius of the circle= Area of the circle = 2 M Find the equation of the sphere that passes through the circle x y z 2x 3y 4z 8 0, x 2y z 8 0 and has its centre on the plane 4x 5y z 3 0. Solution. Equation of a sphere is x y z 2x 3y 4z 8 k(x 2y z 8) 0 Centre= k 3 k ( u, v, w) 1, k, 2 Since the centre lies on the plane 4x 5y z 3 0, we have k 3 k 41 5 k k 9 13 Required equation of a sphere is 13(x y z 2x 3y 4z 8) 9(x 2y z 8) 0.

5 3. Find the equation of the sphere having the circle x y z 10y 4z 8 0, x y z 3 as a great circle. Solution. The equation of the sphere passing through the given circle is x y z 10y 4z 8) k(x y z 3) 0 The given circle is a great circle of this sphere if the centre of the sphere and the centre of the circle coincide. This is possible if the centre of the sphere lies in the plane x y z 3 of the given circle. Center of the sphere is This centre lies on the plane if k 4 Equation of the sphere is x y z 4x 6y 8z 4 0 k k k, 5,2. k k k Exercise. 1.Show that the plane x 2y z 3 cuts the sphere x y z x z 2 0 in a circle of radius unity. Also find the equation of the sphere which has this circle as a great circle. 2. Find the spheres passing through the circle x y z 6x 2z 5 0, y 0 and touching the plane 3y 4z 5 0. Orthogonal spheres: Two spheres are said to be Orthogonal if the tangent planes at a point of intersection are at right angles. The radii of such spheres through their point of intersection P being perpendicular to the tangent planes at P are also at right angles.

6 Figure Thus the spheres cut orthogonally, if the square of the distance between their centers equal to the sum of the squares of their radii. C C r r Show that the condition for the spheres x y z 2u1x 2v1y 2w1z d1 0 and x y z 2u2x 2v2y 2w2z d2 0 to cut orthogonally is 2u1u 2 2v1v2 2w1w2 d1 d2. Solution : The centers of the sphere are C 1 ( u 1, v 1, w 1), C 2 ( u 2, v 2, w 2) and r u v w d, r u v w d 2 Spheres will cut orthogonally if C C r r (u u ) (v v ) (w w ) u v w d u v w d Simplifying, 2u1u 2 2v1v2 2w1w2 d1 d2 is the required condition.. Tangent plane to a sphere: A plane q touches a sphere S if the perpendicular distance of the centre C of S from q is equal to the radius r of S. Then q is called a tangent plane to S. Equation of the sphere S is x y z 2ux 2vy 2wz d 0, C ( u, v, w). Let A (x 1, y 1,z 1) be the point of contact between S and q.

7 Then the direction ratios of CA= (x 1 u, y 1 v,z 1 w). Consider any point P(x,y,z) in the tangent plane. The the direction ratios of AP (x x 1, y y 1,z z 1). Since the radius CA is perpendicular to q, we have (x x 1)(x1 u) (y y 1)(y1 v) (z z 1)(z1 w) 0 xx yy zz u(x x ) v(y y ) w(z z ) x y z Since the point A lies on S, we have x y z 2ux 2vy 2wz d xx 1 yy 1 zz 1 u(x x 1 ) v(y y 1 ) w(z z 1 ) 2ux 1 2vy 1 2wz 1 d xx 1 yy 1 zz 1 u(x x 1 ) v(y y 1 ) w(z z 1 ) d 0 This is the equation of the tangent plane q to S at the point A (x 1, y 1,z 1). Example 1. Find the equation of the tangent plane at a point (1,2, 2) to the sphere x y z 2x 6y 0. Solution. (u, v, w) ( 1, 3,0), d=0 Equation of the tangent plane is x 2y 2z (x 1) 3(y 2) 0 y 2z Find the tangent planes to the sphere x y z 4x 2y 6z 5 0 which are parallel to the plane 2x 2y z 0. Solution. Any plane parallel to the given plane is 2x 2y z k 0 This plane is a tangent plane to the given sphere S if the perpendicular distance p of the centre C of S is equal to the radius r of S. C (2, 1,3), r p k k1 3 p r if k 1 =3 3 k 1 9, k 10 or -8.

8 Equation of the tangent plane are 2x 2y z 10 0 and 2x 2y z 8 0. Right Circular Cone : Definition : A right circular cone is a surface generated by a straight line which passes through a fixed point and makes a constant angle with a fixed line. The constant angle θ is called the semi vertical angle, a fixed point is called a vertex and the fixed line AP is called the axis. Figure: A(vertex) Generator Axis C p B Equation of a right circular cone: Let (x 0, y 0, z 0) be the co-ordinates of the vertex A and (a, b, c) be the direction ratios of the axis. Consider any point P(x, y, z) on the cone. Then the direction ratios of the generator AP are (x x 0, y y 0, z z 0) and cos a(x x ) b(y y ) c(z z ) a (x x ) (y y ) (z z ) Or a(x x ) b(y y ) c(z z ) (a b c ) (x x ) (y y ) (z z ) cos This equation holds for any point P on the cone and hence is the equation of the right circular cone.

9 Example 1: Find the equation of the right circular cone whose vertex is the origin, whose axis is the line x y z and which has semi vertical angle of Solution. Given (x 0, y 0,z 0) (0,0,0). Direction ratios of the axis are (1,2,3) and Then the required equation is (x 2y 3z) (1 2 3 ) x y z cos x 4y 9z 4xy 12yz 6xz 14(x y z ) 4 19x 13y 3z 8xy 24yz 12xz 0. Example 2: Find the equation of the right circular cone generated when the straight line 2y 3z 6, x 0 revolves about z- axis. Solution. Let P(x, y,z) be any point on the cone. The vertex is the point of intersection of the line 2y 3z 6, x 0 and the z- axis. Therefore the vertex A is (0,0,2). A generator of the cone is x y z 2. Direction ratios of the generator are (0,3,2) and the axis are (0,0,1). The semi vertical angle θ is given by cos 2 13 Let P(x,y,z) be any point on the cone so that the direction ratios of AP are (x,y,z-2). Since AP makes an angle θ with AZ, we have x.0 y.0 (z 2).1 cos x y (z 2) 2 4 (z 2) 13 x y (z 2) 4x 2 4y 2 9z 2 36z 36 0, which is the required equation of the cone.

10 Right circular cylinder: Definition: A right circular cylinder is a surface generated by a straight line which is parallel to a straight line and is at a constant distance from it. The constant distance is called the radius of the cylinder. Equation of a right circular cylinder: P L R A Q Let (l,m,n) be the direction cosines of the axis and A(x 0, y 0, z 0 ) be a point on L. Consider an arbitrary point P(x,y,z) on the cylinder. If Q is the foot of the perpendicular from P onto L, then PQ=R, the radius of the cylinder. Also AQ is the projection of AP on L. AQ l(x x 0) m(y y 0) n(z z 0) Also AP (x x ) (y y ) (z z ) Then AP AQ PQ is the required equation of the cylinder. Example 1. The radius of a normal section of a right circular cylinder is 2 units. x 1 y 3 z 2 The axis lies along the straight line. Find its equation Solution.

11 Figure Z R P O A Q Let P(x,y,z) be any point on the cylinder. Draw PN perpendicular to the axis AN. Then PN=2. AN is the projection of AP on AN. Directio ratios of AN are (2,-1,5). Direction cosines of AN= Then the required equation is 2 1 5,, (x 1) (y 3) (z 2) (x 1) (y 3) (z 2) Example 2: Find the equation of the right circular cylinder having the circle x y z 9, x y z 3 as base circle. Solution. The axis of the cylinder is the line through the centre of S and perpendicular to the plane q. We note that O(0,0,0) is the center of S and (1,-1,1) are the direction ratios of the normal to q. The direction cosines of the axis are 1 1 1,,. The perpendicular distance from O on to the plane q is 3 OA If Q is a point common to S and q, then OQ= radius of the sphere= AQ OQ OA 9 3 6

12 Let P(x,y,z) be any point on the cylinder. Then ZP= 6 The equation of the cylinder is (x 0) (y 0) (z 0) (x 0) (y 0) (z 0) x 2 y 2 z 2 xy yz zx 9. 2

CHAPTER 3. Cylinder. ) be any point on the cylinder. The equation of the generator through the point P and parallel to equation (2) are

CHAPTER 3. Cylinder. ) be any point on the cylinder. The equation of the generator through the point P and parallel to equation (2) are CAPTER 3 Cylinder 3.. DENTON A cylinder is a surface generated by a straight line which is parallel to a fixed line and intersects a given curve or touches a given surface. The fixed line is called the