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Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 1
The Sphere
Definition: The set of all points in space that are equidistant from a fixed point is called a sphere. The constant
distance is called the radius of the sphere and the fixed point is called the centre of the sphere.
Example#36: Find the centre and radius of the sphere ๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ โ ๐๐ฑ + ๐๐ฒ โ ๐๐ณ โ ๐๐ = ๐.
Solution: Given equation of sphere is
x2 + y2 + z2 โ 4x + 2y โ 6z โ 11 = 0
x2 โ 4x + y2 + 2y + z2 โ 6z โ 11 = 0
x2 โ 2 2 x + (2)2 + y2 + 2 1 y + 1 2 + z2 โ 2 3 z + 3 2 โ 11 = (2)2 + (1)2 + 3 2
x โ 2 2 + y + 1 2 + z โ 3 2 = 4 + 1 + 9 + 11
x โ 2 2 + y + 1 2 + z โ 3 2 = 25
Which represents the equation of a sphere with centre 2, โ1,3 and radius = 25 = 5
Example#37:Find an equation of the sphere with centre at ๐ ๐, ๐, โ๐ and tangent to the plane
๐๐ฑ โ ๐๐ฒ + ๐๐ณ โ ๐๐ = ๐. Solution:Given equation of the plane 2x โ 3y + 2z โ 10 = 0 and centre of sphere Is at M 4,1, โ6 .
According to given condition , Given plane is tangent to the sphere , So
radius of the sphere = Distance from plane to the Centre
r = 2 4 โ3 1 +2 โ6 โ10
22+ โ3 2+22 โ r =
8โ3โ12โ10
4+9+4 โ r =
โ17
17 โ r =
17
17 โ r = 17
Hence, equation of the sphere with centre 4,1, โ6 and radius 17 will become
x โ 4 2 + y โ 1 2 + z + 6 2 = 17 2
x2 + 16 โ 8x + y2 + 1 โ 2y + z2 + 36 + 12z = 17
x2 + y2 + z2 โ 8x โ 2y + 12z + 36 + 17 = 17
x2 + y2 + z2 โ 8x โ 2y + 12z + 36 = 0 required eq. of sphere
Example#38: Find an equation of the tangent plane to the sphere
๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ โ ๐๐ฑ + ๐๐ฒ โ ๐๐ณ = ๐ at the point ๐, ๐, ๐ . Solution: Given equation of the sphere
x2 + y2 + z2 โ 4x + 2y โ 6z = 0 at the point 3,2,5
Hence Centre of given sphere is โf, โg, โh = 2, โ1,3
Given point on the tangent plane is P 3,2,5
Hence CP is the normal vector of required plane
CP = OP โ OC โ CP = 3,2,5 โ 2, โ1,3 โ CP = i + 3j + 2k
CP = i + 3j + 2k here a = 1, b = 3, c = 2 are direction ratios of normal vector of the plane
Now equation of the tangent at the point 3,2,5 with direction ratios a, b& ๐ of the normal vector of the plane.
a x โ x1 + b y โ y1 + c z โ z1 = 0 โ 1 x โ 3 + 3 y โ 2 + 2 z โ 5 = 0
โ x โ 3 + 3y โ 6 + 2z โ 10 = 0 โ x + 3y + 2z โ 19 = 0 required eq. of the plane
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 2
Question#1: Show that ๐ = ๐ is an equation of a sphere of radius c and Centre at (0,0,0).
Solution: Given ฯ = c
As we know that ฯ = x2 + y2 + z2
โ x2 + y2 + z2 = c
Now squaring on both sides
โ x2 + y2 + z2 = c2
โ (x โ 0)2 + y โ 0 2 + (z โ 0)2 = c2 This equation shows that radius r = c & Centre is (0,0,0)
Question#2: Find an equation of the sphere whose Centre is on the y-axis and which passes through the points
(0,2,2) & (4,0,0).
Solutoin:
Consider a sphere whose Centre C (0,b,0) which passes through the points A(0,2,2) & B (4,0,0) as shown
in the figure.
from figure , we can see that
CA = CB โต radius = CA = CB
(0 โ 0)2 + b โ 2 2 + (0 โ 2)2 = (0 โ 4)2 + b โ 0 2 + (0 โ 0)2
Squaring on both sides
โ 0 + b2 โ 4b + 4 + 4 = 16 + b2 + 0 โ โ4b + 8 = 16 โ 4b = โ8
Thus the co-ordinates of the Centere of sphere are C(0,-2,0)
Now radius = CA = (0 โ 0)2 + (2 โ (โ2))2 + (2 โ 0)2
= (0 โ 0)2 + 2 + 2 2 + 2 โ 0 2
= 0 + 16 + 4
โ radius = 20
Now the equation of sphere having Centere (0,โ2,0) and radius 20
โ (x โ 0)2 + y + 2 2 + z โ 0 2 = 20 2
โ x2 + y2 + 4 + 4y + z2 = 20
โ x2 + y2 + z2 + 4y โ 16 = 0
This is required equation of the sphere
โ ๐ = โ2
Exercise #8.11
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 3
Question#3: Show that an equation of the sphere having the straight line joining the points ๐ฑ๐, ๐ฒ๐, ๐ณ๐ and
๐ฑ๐, ๐ฒ๐, ๐ณ๐ as a diameter is ๐ฑ โ ๐ฑ๐ ๐ฑ โ ๐ฑ๐ + ๐ฒ โ ๐ฒ๐ ๐ฒ โ ๐ฒ๐ + ๐ณ โ ๐ณ๐ ๐ณ โ ๐ณ๐ = ๐. Solution:Consider a sphere having the straight line joining the points A x1 , y1 , z1 and B x2 , y2 , z2 as diameter.
Consider a point P on the sphere and meet A and B.
From figure AP is perpendicular to BP
Now direction ratios of line AP are x โ x1 , y โ y1 , z โ z1
& ๐๐๐๐๐๐ก๐๐๐ ๐๐๐ก๐๐๐ ๐๐ ๐๐๐๐ ๐ต๐ ๐๐๐ ๐ฅ โ x2 , y โ y2 , z โ z2
Therefore AP = x โ x1 i + y โ y1 j + z โ z1 k & BP = x โ x2 i + y โ y2 j + z โ z2 k
From figure AP โฅ BP then AP . BP = 0
โ x โ x1 i + y โ y1 j + z โ z1 k . x โ x2 i + y โ y2 j + z โ z2 k = 0
โ x โ x1 x โ x2 + y โ y1 y โ y2 + z โ z1 z โ z2 = 0 is required equation of sphere.
Question#4: Find an equation of the sphere which passes through the points ๐ โ๐, ๐, ๐ , ๐ โ๐, โ๐, โ๐
๐๐ง๐ ๐ ๐, ๐, ๐ and whose centre lies on the hypotenuse of the right triangle ๐๐๐. Solution: Given points are A โ3,6,0 , B โ2, โ5, โ1 & ๐ถ 1,4,2
For required proof wehve to find
AB = (โ2 + 3)2 + (โ5 โ 6)2 + (โ1 โ 0)2 = (1)2 + โ11 2 + โ1 2 โ AB = 123
BC = (1 + 2)2 + 4 + 5 2 + 2 + 1 2 = (3)2 + 9 2 + 3 2 โ BC = 99
CA = (โ3 โ 1)2 + (6 โ 4)2 + (0 โ 2)2 = (โ4)2 + 2 2 + โ2 2 โ CA = 24
From above equations its clear that AB 2
= BC 2
+ CA 2
So by Pethagoras theorem , AB is the hypotenuse of โABC.
So AB act as a diameter of sphere.The mid point of AB is M called Centre of sphere.
M = โ2โ3
2,โ5+6
2,
0โ1
2 =
โ5
2,
1
2,โ1
2
Now radius of sphere = AM = โ5
2+ 3
2+
1
2โ 6
2+
โ1
2โ 0
2 =
1
4+
121
4+
1
4 =
123
4
Hence equation of sphere with Centre M = โ5
2,
1
2,โ1
2 radius= AM =
123
4.
x โ โ5
2
2
+ y โ1
2
2
+ z โ โ1
2
2
= 123
4
2
โ x +5
2
2
+ y โ1
2
2
+ z +1
2
2
=123
4
x2 +25
4+ 5x + y2 +
1
4โ y + z2 +
1
4+ z =
123
4
x2 + y2 + z2 + 5x โ y + z =123
4โ
25
4โ
1
4โ
1
4
x2 + y2 + z2 + 5x โ y + z =94
4 โ x2 + y2 + z2 + 5x โ y + z = 24
โ x2 + y2 + z2 + 5x โ y + z โ 24 = 0 ๐ซ๐๐ช๐ฎ๐ข๐ซ๐๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐จ๐ ๐ฌ๐ฉ๐ก๐๐ซ๐.
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 4
Question#5: Prove that each of the following equation represents a sphere. Find the centre and radius of each:
(i) ๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ โ ๐๐ฑ + ๐๐ณ = ๐
Solution: Given equation of sphere is
x2 + y2 + z2 โ 6x + 4z = 0
x2 โ 6x + y2 + z2 + 4z = 0
x2 โ 2 x 3 + (3)2 + (y โ 0)2 + z2 + 2 z 2 + (2)2 = (3)2 + (2)2
x โ 3 2 + y โ 0 2 + z + 2 2 = 13
x โ 3 2 + y โ 0 2 + z โ (โ2) 2 = 13 2
Which represents the equation of a sphere with centre 3,0, โ2 and radius 13.
(ii) ๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ + ๐๐ฑ โ ๐๐ฒ โ ๐๐ณ + ๐ = ๐
Solution: Given equation of sphere is
x2 + y2 + z2 + 2x โ 4y โ 6z + 5 = 0
x2 + 2x + y2 โ 4y + z2 โ 6z + 5 = 0
x2 + 2 x 1 + 1 2 + y2 โ 2 y 2 + 2 2 + z2 โ 2 z 3 + 3 2 + 5 = 1 2 + 2 2 + 3 2
(x + 1)2 + (y โ 2)2 + (z โ 3)2 = 1 + 4 + 9 โ 5
(x โ (โ1))2 + (y โ 2)2 + (z โ 3)2 = 9
(x โ (โ1))2 + (y โ 2)2 + (z โ 3)2 = 3 2
Which represents the equation of a sphere with centre โ1,2,3 and radius 3 .
(iii) ๐๐ฑ๐ + ๐๐ฒ๐ + ๐๐ณ๐ โ ๐๐ฑ + ๐๐ฒ + ๐๐๐ณ + ๐ = ๐
Solution: Given equation of sphere is
4x2 + 4y2 + 4z2 โ 4x + 8y + 24z + 1 = 0
Now dividing both sides by 4
x2 + y2 + z2 โ x + 2y + 6z +1
4= 0
x2 โ x + y2 + 2y + z2 + 6z +1
4= 0
x2 โ 2 x 1
2 +
1
2
2
+ y2 + 2 y 1 + (1)2 + z2 + 2 z 3 + 3 2 +1
4=
1
2
2
+ (1)2 + 3 2
x โ1
2
2
+ y + 1 2 + z + 3 2 =1
4+ 1 + 9 โ
1
4
x โ1
2
2+ y โ โ1
2+ z โ (โ3 2 = 10
x โ1
2
2+ y โ โ1
2+ z โ (โ3 2 = 10
2
Which represents the equation of a sphere with centre 1
2, โ1, โ3 and radius 10.
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 5
Question#6: find an equation of the sphere through the points ๐, ๐, ๐ , ๐, ๐, โ๐ , (โ๐, ๐, ๐) and ๐, ๐, ๐ . Also
find its centre and radius.
Solution: Let equation of the sphere is x2 + y2 + z2 + 2fx + 2gy + 2hz + c = 0
It passes through the points
0,0,0 ; 0 + 0 + 0 + 0 + 0 + 0 + c = 0 โ ๐ = ๐
0,1, โ1 ; 0 + 1 + 1 + 0 + 2g + 2h โ1 + 0 = 0 โ 2g โ 2h = โ2 โ g โ h = โ1 โ โ โ 1
โ1,2,0 ; 1 + 4 + 0 โ 2f + 4g + 0 + 0 = 0 โ โ2f + 4g = โ5 โ โ โ 2
1,2,3 ; 1 + 4 + 9 + 2f + 4g + 6h + 0 = 0 โ 2f + 4g + 6h = โ14 โ f + 2g + 3h = โ7 โ โ โ 3
From equation (1) โ g = h โ 1 โ โ โ 4
Now putting the value of g in equation (2)
โ2f + 4 h โ 1 = โ5 โ โ2f + 4h โ 4 = โ5 โ 4h + 1 = 2f โ f =1
2 4h + 1 โ โ โ 5
Now using equations (4) and (5) in equation (3)
1
2 4h + 1 + 2 h โ 1 + 3h = โ7 โ 2h +
1
2+ 2h โ 2 + 3h = โ7
7h = โ7 โ1
2+ 2 โ 7h = โ7 +
3
2 โ 7h =
โ11
2 โ ๐ก =
โ๐๐
๐๐
Putting the value of h in eq (5)
f =1
2 4
โ11
14 + 1 โ f =
1
2
โ44
14+ 1 โ f =
1
2
โ30
14 โ ๐ = โ
๐๐
๐๐
Putting value of h in eq . (1)
โ gโ11
14= โ1 โ g +
11
14= โ1 โ g = โ1 โ
11
14 โ ๐ = โ
๐๐
๐๐
Put values of f , g ,h in eq, (A)
x2 + y2 + z2 + 2 โ15
14 x + 2 โ
25
14 y + 2 โ
11
14 z + 0 = 0
14x2 + 14y2 + 14z2 โ 30x โ 50y โ 22z = 0
7x2 + 7y2 + 7z2 โ 15x โ 25y โ 11z = 0 is required equation
Dividing both sides by 7
x2 + y2 + z2 โ15
7x โ
25
7y โ
11
7z = 0
Its centre is โ15
7(โ2) , โ
25
7 โ2 , โ
11
7 โ2
centre = 15
14 ,
25
14 ,
11
14
Now radius is = 15
14
2+
25
14
2+
11
14
2โ 0 =
225
196+
625
196+
121
196=
225+625+121
196
radius = 971
14
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 6
Question#7: Find an equation of the sphere passing through the points ๐, โ๐, โ๐ , (๐, โ๐, โ๐) and having its
centre on the straight line ๐๐ฑ โ ๐๐ฒ = ๐ = ๐๐ฒ + ๐๐ณ. Solution: Let the general equation of sphere is
x2 + y2 + z2 + 2fx + 2gy + 2hz + c = 0 โ โ โ A
Here (โf, โg, โh) is a Centre of equation (A)
The Centre of the sphere passing through this line
2x โ 3y = 0 , 5y + 2z = 0
2x โ 3y = 0 โ 2 โf โ 3 โg = 0 โ2f + 3g = 0 โ โ โ 1
5y + 2z = 0 โ 5 โg + 2 โh = 0 โ โ5g โ 2h = 0 โ 5g + 2h = 0 โ โ โ 2
Now equation of the sphere passing through the points 0, โ2, โ4 , (2, โ1, โ1).
0, โ2, โ4 ; 0 + 4 + 16 + 0 โ 4g โ 8h + c = 0 โ โ4g โ 8h + c = โ20 โ โ โ 3
2, โ1, โ1 ; 4 + 1 + 1 + 4f โ 2g โ 2h + c = 0 โ 4f โ 2g โ 2h + c = โ6 โ โ โ 4
Now subtracting eq(3) & eq(4) โ4f โ 2g โ 6h = โ14
Dividing by โโโ2โโ above 2f + g + 3h = 7 โ โ โ โ 5
Now using eq(1) & eq(2)
from 1 3g = 2f โ f =3
2g โ โ โ 6
from 2 2h = โ5g โ h =โ5
2g โ โ โ 7
Using equ(6) & equ(7) in equation (5)
2 3
2g + g + 3
โ5
2g = 7 โ 3g + g โ
15
2g = 7 โ 4g โ
15
2= 7 โ g
8 โ 15
2 = 7
g โ7
2 = 7 โ ๐ = โ๐
eq 6 becomes f =3
2 โ2 โ ๐ = ๐
eq 7 becomes h =โ5
2 โ2 =
10
2 โ ๐ก = ๐
Putting these values in eq(3)
โ4 โ2 โ 8 5 + c = โ20 โ c = โ20 + 40 โ 8 โ ๐ = ๐๐
Now equation (A) becomes
x2 + y2 + z2 + 2 โ3 x + 2 โ2 y + 2 5 z + +12 = 0
x2 + y2 + z2 โ 6x โ 4y + 10z + +12 = 0
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 7
Question#8: Find an equation of the sphere which passes through the circle
๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ = ๐, ๐๐ฑ + ๐๐ฒ + ๐๐ณ = ๐, and the point ๐, ๐, ๐
๐๐ข๐ง๐ญ: ๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ โ ๐ + ๐ค ๐๐ฑ + ๐๐ฒ + ๐๐ณ โ ๐ = ๐ ๐๐๐๐ข๐ง๐๐ฌ ๐ ๐ฌ๐ฉ๐ก๐๐ซ๐ ๐ญ๐ก๐ซ๐จ๐ฎ๐ ๐ก ๐ญ๐ก๐ ๐๐ข๐ซ๐๐ฅ๐ ๐๐จ๐ซ ๐๐๐๐ก ๐ค๐๐.
Solution: Let equation of the sphere passes through the circle is
x2 + y2 + z2 โ 9 + k 2x + 3y + 4z โ 5 = 0 โ โ โ 1
This equation of the sphere passes through the point 1,2,3 .
1 + 4 + 9 โ 9 + k 2 + 6 + 12 โ 5 = 0 โ 5 + 15k = 0 โ 5 = โ15k โ ๐ค = โ๐
๐
Using the value of k in equation (1)
x2 + y2 + z2 โ 9 โ1
3 2x + 3y + 4z โ 5 = 0
3x2 + 3y2 + 3z2 โ 27 โ 2x โ 3y โ 4z + 5 = 0
3x2 + 3y2 + 3z2 โ 2x โ 3y โ 4z โ 22 = 0
Question#9: Find an equation of the sphere through the circle ๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ = ๐, ๐๐ฑ + ๐๐ฒ + ๐๐ณ โ ๐ = ๐ and
touching the plane ๐ณ = ๐. Solution: Let equation of the sphere passes through the circle is
x2 + y2 + z2 โ 1 + k 2x + 4y + 5z โ 6 = 0 โ โ โ 1
x2 + y2 + z2 โ 1 + 2kx + 4ky + 5kz โ 6k = 0
x2 + y2 + z2 + 2kx + 4ky + 5kz + โ 1 + 6k = 0
Here the Centre of this equation is โ k, โ2k, โ5
2k & center of the g. eq is โ f, โg, โh
And radius is r = f 2 + g2 + h2 โ c
r = โk 2 + โ2k 2 + โ5
2k
2โ 1 + 6k = k2 + 4k2 +
25k
4
2+ 1 + 6k =
4k2+16k2+25k2+4+24k
4
r = 45k2 + 24k + 4
4
As the sphere touches the plane z = 0 , So Radius of the sphere = Distance between plane & Centre
45k2+24k+4
4=
0+0โ5k
2
02+02+12 โ
45k2+24k+4
4=
โ5k
2
1 โ
45k2+24k+4
4=
5k
2
Now squaring on both sides
45k2+24k+4
4=
25k2
4 โ 45k2 + 24k + 4 = 25k2 โ 45k2 โ 25k2 + 24k + 4 = 0 โ 20k2 + 24k + 4 = 0
dividing by 4 โ 5k2 + 6k + 1 = 0 โ 5k2 + 6k + 1 = 0 โ 5k2 + 5k + k + 1 = 0
5k k + 1 + k + 1 = 0 โ 5k + 1 k + 1 = 0 โ 5k + 1 = 0, k + 1 = 0 โ ๐ค = โ๐, โ๐
๐
Put k = โ1, โ1
5in Equation (1)
x2 + y2 + z2 โ 1 โ 1 2x + 4y + 5z โ 6 = 0 x2 + y2 + z2 โ 1 โ1
5 2x + 4y + 5z โ 6 = 0
x2 + y2 + z2 โ 1 โ 2x โ 4y โ 5z + 6 = 0 5x2 + 5y2 + 5z2 โ 5 โ 2x โ 4y โ 5z + 6 = 0
x2 + y2 + z2 โ 2x โ 4y โ 5z + 5 = 0 5x2 + 5y2 + 5z2 โ 2x โ 4y โ 5z + 1 = 0
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 8
Question#10: Show that the two circles ๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ = ๐, ๐ฑ โ ๐๐ฒ + ๐๐ณ โ ๐๐ = ๐ and ๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ + ๐๐ฒ โ ๐๐ณ +๐๐ = ๐, ๐ฑ + ๐ฒ + ๐ณ + ๐ = ๐ lie on the same sphere. Also find its equation.
Solution: Let equation of the sphere passes through the first circle is
x2 + y2 + z2 โ 9 + k x โ 2y + 4z โ 13 = 0 โ โ โ A
x2 + y2 + z2 + kx โ 2ky + 4kz โ 9 + 13k = 0 โ โ โ (1)
x2 + y2 + z2 + 6y โ 6z + 21 + h x + y + z + 2 = 0 โ โ โ (B)
x2 + y2 + z2 + hx + h + 6 y + h โ 6 z + 2h + 21 = 0 โ โ(2)
The given circles lie on the same sphere then eq.(1) & eq.(2) must be identical.
Comparing co-efficients of x, y, z & constants in (1) & (2)
k = h โ โ โ i
โ2k = h + 6 โ โ โ ii
4k = h โ 6 โ โ โ iii
โ 9 + 13k = 2h + 21 โ โ โ (iv)
From i k = h put in ii
โ2h = h + 6 โ โ3h = 6 โ ๐ก = โ๐ & also ๐ค = โ๐
Putting these values in (iii) & (iv)
We see that eqs. (iii) & (iv) are satisfied.
Values of k & h shows that the both circles lie on the same sphere and equation of the sphere is obtained by putting
k = โ2 in (A) or in (B).
x2 + y2 + z2 โ 9 โ 2 x โ 2y + 4z โ 13 = 0
x2 + y2 + z2 โ 9 โ 2x + 4y โ 8z + 26 = 0
x2 + y2 + z2 โ 2x + 4y โ 8z + 17 = 0
Question#11: Find an equation of the sphere for which the circle ๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ + ๐๐ฒ โ ๐๐ณ + ๐ = ๐ , ๐๐ฑ + ๐๐ฒ โ ๐๐ณ โ ๐ = ๐ is a great circle.
๐๐ข๐ง๐ญ โถ ๐ ๐ ๐ซ๐๐๐ญ ๐๐ข๐ซ๐๐ฅ๐ ๐จ๐ ๐ฌ๐ฉ๐ก๐๐ซ๐ ๐ข๐ฌ ๐ฌ๐ฎ๐๐ก ๐ญ๐ก๐๐ญ ๐ข๐ญ๐ฌ ๐ฉ๐ฅ๐๐ง๐ ๐ฉ๐๐ฌ๐ฌ๐๐ฌ ๐ญ๐ก๐ซ๐จ๐ฎ๐ ๐ก ๐ญ๐ก๐ ๐๐๐ง๐ญ๐ซ๐ ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ฉ๐ก๐๐ซ๐ Solution: Let equation of the sphere passes through a given circle is
x2 + y2 + z2 + 7y โ 2z + 2 + k 2x + 3y + 4z โ 8 = 0 โ โ โ 1
x2 + y2 + z2 + 7y โ 2z + 2 + 2kx + 3ky + 4kz โ 8k = 0
x2 + y2 + z2 + 2kx + 3k + 7 y + 4k โ 2 z + 2 โ 8k = 0
Now Centre of this sphere is โ f, โg, โh = โ k, โ3k+7
2, 1 โ 2k
According to given condition that the given circle is a great circle.
Then we know that plane of the circle passes through Centre of the sphere.
Then 2 โk + 3 โ3k+7
2 + 4 1 โ 2k โ 8 = 0
โ โ2k โ3
2 3k + 7 + 4 1 โ 2k โ 8 = 0
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
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โ โ4k โ 3 3k + 7 + 8 โ 16k โ 16 = 0
โ โ4k โ 9k โ 21 + 8 โ 16k โ 16 = 0
โ โ29k โ 29 = 0 โ k + 1 = 0 โ ๐ค = โ๐
Using value of k in eq.(1)
x2 + y2 + z2 + 7y โ 2z + 2 โ 1 2x + 3y + 4z โ 8 = 0
x2 + y2 + z2 + 7y โ 2z + 2 โ 2x โ 3y โ 4z + 8 = 0
x2 + y2 + z2 โ 2x + 4y โ 6z + 10 = 0 required equation pf sphere
Question#12: Find an equation of the sphere with centre ๐, โ๐, โ๐ and tangent to the plane ๐ฑ โ ๐๐ฒ + ๐ณ + ๐ = ๐.
Solution:
Given equation of the plane x โ 2y + z + 7 = 0 and centre of sphere is 2, โ1, โ1 .
According to given condition , Given plane is tangent to the sphere
So radius of the sphere = Distance from plane to the Centre
r = 2โ2 โ1 + โ1 +7
12+ โ2 2+12
r = 2+2โ1+7
1+4+1
r = 10
6
r =10
6
Hence equation of the sphere with centre 2, โ1, โ1 and radius 10
6 will become
x โ 2 2 + y + 1 2 + z + 1 2 = 10
6
2
x2 + 4 โ 4x + y2 + 1 + 2y + z2 + 1 + 2z =100
6
x2 + y2 + z2 โ 4x + 2y + 2z + 6 =50
3
3 x2 + y2 + z2 โ 4x + 2y + 2z + 6 = 50
3x2 + 3y2 + 3z2 โ 12x + 6y + 6z + 18 = 50
3x2 + 3y2 + 3z2 โ 12x + 6y + 6z โ 32 = 0 is required eq. of sphere
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
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Question#13: Find an equation of the plane tangent to the sphere
๐ ๐ฑ๐ + ๐ฒ๐ + ๐ณ๐ โ ๐๐ฑ โ ๐๐ฒ โ ๐๐ณ โ ๐๐ = ๐ at the point ๐, ๐, ๐ .
Solution: Given equation of the sphere
3 x2 + y2 + z2 โ 2x โ 3y โ 4z โ 22 = 0 at the point 1,2,3
Dividing by 3 on both sides
x2 + y2 + z2 โ2
3x โ y โ
4
3z โ
22
3= 0
Hence Centre of given sphere is โf, โg, โh = 1
3,
1
2,
2
3
Given points on the tangent plane are P 1,2,3
Hence CP is the normal vector of required plane
CP = OP โ OC โ CP = 1,2,3 โ 1
3,1
2,2
3
CP = 2
3i +
3
2j +
7
3k Here a =
2
3, b =
3
2, c =
7
3 are direction ratios of normal vector of the plane
Now equation of the tangent at the point 1,2,3 with direction ratios a, b& c of the normal vector of the plane.
a x โ x1 + b y โ y1 + c z โ z1 = 0
2
3 x โ 1 +
3
2 y โ 2 +
7
3 z โ 3 = 0
Multiplying both sides by 6
4 x โ 1 + 9 y โ 2 + 14 z โ 3 = 0
4x โ 4 + 9y โ 18 + 14z โ 42 = 0
4x + 9y + 14z โ 64 = 0 required eq. of the plane
Question#14: Find an equation of the sphere with centre at the point (โ๐, ๐, โ๐) and tangent to the
๐ ๐ฑ๐ฒ โ ๐ฉ๐ฅ๐๐ง๐ (๐) ๐ฒ๐ณ โ ๐ฉ๐ฅ๐๐ง๐ (๐)๐ณ๐ฑ โ ๐ฉ๐ฅ๐๐ง๐ Solution (a):
Equation of xy โ plane is z = 0
As xy โ plane is tangent to the required sphere with centre at the point P(โ2,4, โ6)
So radius of the sphere = Distance between centre of sphere and plane
radius = 0x + 0y + โ6 z + 0
0 + 0 + 1
radius = โ6
1 โ radius = 6
Now equation of the sphere with centre (โ2,4, โ6) and radius 6 will become
x + 2 2 + y โ 4 2 + z + 6 2 = 6 2
x2 + 4 + 4x + y2 + 16 โ 8y + z2 + 36 + 12z = 36
x2 + y2 + z2 + 4x โ 8y + 12z + 20 = 0
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
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Solution(b): Equation of yz โ plane is x = 0
As yz โ plane is tangent to the required sphere with centre at the point P(โ2,4, โ6)
So radius of the sphere = Distance between centre of sphere and plane
radius = (โ2)x+0y+0z+0
1+0+0 โ radius =
โ2
1 โ radius = 2
Now equation of the sphere with centre (โ2,4, โ6) and radius 2 will become
x + 2 2 + y โ 4 2 + z + 6 2 = 2 2
x2 + 4 + 4x + y2 + 16 โ 8y + z2 + 36 + 12z = 4
x2 + y2 + z2 + 4x โ 8y + 12z + 52 = 0
Solution(c): Equation of zx โ plane is y = 0
As zx โ plane is tangent to the required sphere with centre at the point P(โ2,4, โ6)
So radius of the sphere = Distance between centre of sphere and plane
radius = 0x+(4)y+0z+0
0+1+0 โ radius =
4
1 โ radius = 4
Now equation of the sphere with centre (โ2,4, โ6) and radius 4 will become
x + 2 2 + y โ 4 2 + z + 6 2 = 4 2
x2 + 4 + 4x + y2 + 16 โ 8y + z2 + 36 + 12z = 16
x2 + y2 + z2 + 4x โ 8y + 12z + 40 = 0
Question#15: Find an equation of the surface whose points are equidistant from ๐(๐, ๐, ๐) and ๐ ๐, ๐, โ๐ .
Solution: Given points are P 7,8,2 and Q 5,2, โ6
Let R(x, y, z) be a point on the required surface then by given condition
RP = PQ
7 โ x 2 + 8 โ y 2 + 2 โ z 2 = 5 โ x 2 + 2 โ y 2 + โ6 โ z 2
Taking square on both sides
7 โ x 2 + 8 โ y 2 + 2 โ z 2 = 5 โ x 2 + 2 โ y 2 + โ6 โ z 2
7 โ x 2 + 8 โ y 2 + 2 โ z 2 = 5 โ x 2 + 2 โ y 2 + โ 6 + z 2
7 โ x 2 + 8 โ y 2 + 2 โ z 2 = 5 โ x 2 + 2 โ y 2 + 6 + z 2
49 โ 14x + x2 + 64 โ 16y + y2 + 4 โ 4z + z2 = 25 + x2 โ 10x + 4 โ 4y + y2 + 36 + z2 + 12z
117 โ 14x โ 16y โ 4z โ 65 + 10x + 4y โ 12z = 0
โ โ4x โ 12y โ 16z + 52 = 0
โ โ4 x + 3y + 4z + 13 = 0
โ x + 3y + 4z + 13 = 0
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
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Question#16: A point ๐ moves such that the square of its distance from the origin is proportional to its distance
from a fixed plane. Show that P always lie on a sphere.
Solution: Let equation of the fixed plane is ๐๐ฅ + ๐๐ฆ + ๐๐ง = ๐ โ โ โ 1
Where ๐, ๐ &๐ are the direction cosines of the normal vector of the plane (1)
Let ๐(๐ฅ, ๐ฆ, ๐ง) be any point on the locus then according to given condition
๐๐ 2
โ ๐๐
๐๐ 2
= ๐ ๐๐ where k is constant.
๐ฅ โ 0 2 + ๐ฆ โ 0 2 + ๐ง โ 0 2 = ๐ ๐๐ฅ + ๐๐ฆ + ๐๐ง โ ๐
๐2 + ๐2 + ๐2
๐ฅ2 + ๐ฆ2 + ๐ง2 = ๐ ๐๐ฅ + ๐๐ฆ + ๐๐ง โ ๐ โต ๐2 + ๐2 + ๐2 = 1
๐ฅ2 + ๐ฆ2 + ๐ง2 โ ๐ ๐๐ฅ + ๐๐ฆ + ๐๐ง โ ๐ = 0
๐ฅ2 + ๐ฆ2 + ๐ง2 โ ๐๐๐ฅ โ ๐๐๐ฆ โ ๐๐๐ง + ๐๐ = 0
Which is clearly the equation of a sphere .
Hence proved that the point ๐ always lies on a sphere.
Question#17: A sphere of radius ๐ passes through the origin and meets the axes in ๐จ, ๐ฉ, ๐ช. Prove that the
centroid of the triangle ๐จ๐ฉ๐ช lies on the sphere ๐ ๐๐ + ๐๐ + ๐๐ = ๐๐๐.
Solution: Let equation of the sphere passes through the origin is
๐ฅ2 + ๐ฆ2 + ๐ง2 + 2๐๐ฅ + 2๐๐ฆ + 2๐๐ง = 0
Centre of sphere is โ ๐, โ๐, โ๐ . & radius is ๐ = ๐2 + ๐2 + ๐2
Given radius ๐ = ๐ ๐ก๐๐๐ ๐๐๐๐ฃ๐ ๐๐ฅ๐๐๐๐ ๐ ๐๐๐ ๐ค๐๐๐ ๐๐๐๐๐๐๐
๐ = ๐2 + ๐2 + ๐2 โ ๐2 = ๐2 + ๐2 + ๐2 โ โ โ 1
As the sphere cuts axes in pointโs ๐ด, ๐ต, ๐ถ.So the co-ordinates of the points ๐ด, ๐ต& ๐ถ are
๐ด โ2๐, 0,0 , ๐ต 0, โ2๐, 0 & ๐ถ 0,0, โ2๐ .
As we know that centroid of โ๐ด๐ต๐ถ is ๐ฅ1+๐ฅ2+๐ฅ3
3,๐ฆ1+๐ฆ2+๐ฆ3
3,๐ง1+๐ง+๐ง3
3 .
Now Centroid of โ๐ด๐ต๐ถ is โ2๐
3,โ2๐
3,โ2๐
3 .
Here ๐ฅ =โ2๐
3 , ๐ฆ =
โ2๐
3 , ๐ง =
โ2๐
3
Putting in given equation 9 ๐ฅ2 + ๐ฆ2 + ๐ง2 = 4๐2
9 4๐2
9+
4๐2
9+
4๐2
9 = 4๐2
9
9 4๐2 + 4๐2 + 4๐2 = 4๐2
4 ๐2 + ๐2 + ๐2 = 4๐2
โ 4๐2 = 4๐2 ๐๐๐๐ ๐๐. (1) As this equation is satisfied
So the centroid of the triangle ๐ด๐ต๐ถ lies on the sphere 9 ๐ฅ2 + ๐ฆ2 + ๐ง2 = 4๐2.
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 13
Question#18: The plane ๐
๐+
๐
๐+
๐
๐= ๐meets the axes in ๐จ, ๐ฉ, ๐ช. Find an equation of the circumcircle of the
triangle ๐จ๐ฉ๐ช. Also find the co-ordinates of the centre of the circle.
Solution: The required circle is the intersection of the given plane by a sphere through origin O and points ๐ด, ๐ต & ๐ถNow given equation of the plane is
๐ฅ
๐+
๐ฆ
๐+
๐ง
๐= 1 โ โ โ 1
As this plane cuts co-ordinate axes in points A.B & C.
So co-ordinates of these points are ๐ด ๐, 0,0 , ๐ต 0, ๐, 0 , ๐ถ 0,0, ๐ & ๐ 0,0,0
Now equation of the sphere OABC is
๐ฅ2 + ๐ฆ2 + ๐ง2 โ ๐๐ฅ โ ๐๐ฆ โ ๐๐ง = 0 โ โ โ 2
The eq.(1) and eq.(2) are the equations of required circumcircle
Now centre of the circle is the foot of perpendicular from the centre of sphere (1) to plane (2).
Here centre of sphere (2) is ๐บ ๐
2,๐
2,๐
2
Now direction ratios of line perpendicular to plane (2) are 1
๐,
1
๐,
1
๐
So equation of line PG is
๐ฅโ
๐
21
๐ =
๐ฆโ ๐
21
๐ =
๐งโ ๐
21
๐ = ๐ก
๐ฅ =๐
2+
๐ก
๐
๐ฆ =๐
2+
๐ก
๐
๐ง =๐
2+
๐ก
๐
So co-ordinates of point P are ๐ ๐
2+
๐ก
๐,๐
2+
๐ก
๐,๐
2+
๐ก
๐
As the point P lies on the plane (2)
So 1
๐
๐
2+
๐ก
๐ +
1
๐
๐
2+
๐ก
๐ +
1
๐
๐
2+
๐ก
๐ = 1
1
2+
๐ก
๐2+
1
2+
๐ก
๐2+
1
2+
๐ก
๐2= 1
3
2+ ๐ก
1
๐2+
1
๐2+
1
๐2 = 1
๐ก 1
๐2+
1
๐2+
1
๐2 = โ1
2
๐ก =โ1
2 1
๐2+1
๐2+1
๐2
By putting this value of t in eq.(A) we get co-ordinates of centre of circle.
Mathematics Calculus With Analytic Geometry by SM. Yusaf & Prof.Muhammad Amin
Written and composed by M.Bilal (4363500@gmail.com) Mardawal Naushehra ,KHUSHAB Page 14
Question#19: A plane passes through a fixed point (๐, ๐, ๐) and cuts the axes of coordinates in ๐จ, ๐ฉ , ๐ช.Find the
locus of the centre of the sphere ๐ถ๐จ๐ฉ๐ช for different positons of the planes, O being the origin.
Solution: Let equation of the fixed plane is ๐๐ฅ + ๐๐ฆ + ๐๐ง = ๐ โ โ โ 1
As this plane passes through a point (๐, ๐, ๐) , so ๐๐ + ๐๐ + ๐๐ = ๐ โ โ โ 2
Equation(1) cuts at ๐ด, ๐ต & ๐ถ Then ๐ด โ๐
๐, 0,0 , ๐ต 0, โ
๐
๐, 0 & ๐ถ 0,0, โ
๐
๐
Thus equation of sphere OABC will be ๐ฅ2 + ๐ฆ2 + ๐ง2 โ๐
๐๐ฅ โ
๐
๐๐ฆ โ
๐
๐๐ง = 0
Hence centre of sphere will be โ๐, โ๐, โ๐ = ๐
2๐,
๐
2๐,
๐
2๐
Let ๐ฅ1 =๐
2๐ , ๐ฆ1 =
๐
2๐ , ๐ง1 =
๐
2๐
โ ๐ =๐
2๐ฅ1 , ๐ =
๐
2๐ฆ1 , ๐ =
๐
2๐ง1
Using these values in ๐๐. 2
โ๐
2๐ฅ1๐ +
๐
2๐ฆ1๐ +
๐
2๐ง1๐ = ๐ โ
๐
2๐ฅ1+
๐
2๐ฆ1+
๐
2๐ง1= 1 โ
1
2
๐
๐ฅ1+
๐
๐ฆ1+
๐
๐ง1 = 1 โ
๐
๐ฅ1+
๐
๐ฆ1+
๐
๐ง1= 2 This is the required
equation.
Question#20: Find an equation of the sphere circumscribing the tetrahedron whose faces are
๐ = ๐, ๐ = ๐, ๐ = ๐ and ๐๐ + ๐๐ + ๐๐ + ๐ = ๐. Solution: Let equation of the sphere is
๐ฅ2 + ๐ฆ2 + ๐ง2 + 2๐๐ฅ + 2๐๐ฆ + 2๐๐ง + ๐ = 0 โ โ โ (1)
Given equation of the plane is ๐๐ฅ + ๐๐ฆ + ๐๐ง + ๐ = 0
Let equation of the sphere passes through the vertices of tetrahedron OABC.
๐ 0,0,0 ; ๐ = 0
๐ด โ๐
๐, 0,0 :
๐2
๐2 โ2๐๐
๐= 0 โ
๐
๐
๐
๐โ 2๐ โ
๐
๐= 2๐ โ ๐ =
๐
2๐
๐ต 0,โ๐
๐, 0 :
๐2
๐2 โ2๐๐
๐= 0 โ
๐
๐
๐
๐โ 2๐ โ
๐
๐= 2๐ โ g =
p
2m
C 0,0,โp
n :
p2
n2 โ2hp
n= 0 โ
p
n
p
nโ 2h โ
p
n= 2h โ h =
p
2n
Using In equation (1), we get x2 + y2 + z2 +p
lx +
p
my +
p
nz = 0
Checked by: Sir Hameed ullah ( hameedmath2017 @ gmail.com)
Specially thanks to my Respected Teachers
Prof. Muhammad Ali Malik (M.phill physics and Publisher of www.Houseofphy.blogspot.com)
Muhammad Umar Asghar sb (MSc Mathematics)
Hameed Ullah sb ( MSc Mathematics)
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