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Edurev123 
4. Sphere and its Properties 
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =?? 
and passing through the ci zle 
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =?? 
(2009 : 12 Marks) 
Solution: 
Approach: General equation of sphere through any circle is used. The parameter san be 
found by the Centre satisfying equation of plane. 
General equation of a sphere passing through the circle is 
?? +???? =0 
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0 
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0 
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if 
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3 
?                                                                                                        3?? +37=6 
?                                                                                                                     ?? =
-31
3
 
? Required sphere is 
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0 
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a 
circle of radius 1 and find the equation of the sphere which has this circle as great 
circle. 
(2010: 12 Marks) 
Solution: 
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3 
Page 2


Edurev123 
4. Sphere and its Properties 
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =?? 
and passing through the ci zle 
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =?? 
(2009 : 12 Marks) 
Solution: 
Approach: General equation of sphere through any circle is used. The parameter san be 
found by the Centre satisfying equation of plane. 
General equation of a sphere passing through the circle is 
?? +???? =0 
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0 
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0 
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if 
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3 
?                                                                                                        3?? +37=6 
?                                                                                                                     ?? =
-31
3
 
? Required sphere is 
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0 
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a 
circle of radius 1 and find the equation of the sphere which has this circle as great 
circle. 
(2010: 12 Marks) 
Solution: 
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3 
 Centre of given circle  =(
1
2
,-
1
2
,0)
 Radius  =
v
1
4
,
1
4
,2=
v
2+
1
2
 =
v
5
2
 
Let ?? be the centre of this circle. 
Distance of plane from centre 
|
1
2
-
1
2
-3|
v1
2
+1
2
+2
2
=
3
v6
 
? Radius of circle with ???? as radius 
v
5
2
-
9
6
=
v
2
2
=1 
 
 
Equation of sphere with circle as great circle. 
                        ?? 2
+?? 2
+?? 2
-?? +?? -2+?? (?? +?? -2?? -3)=0
?                     ?? 2
+?? 2
+?? 2
-?? +?? -2+???? +???? -2???? -?? 3=0
?              ?? 2
+?? 2
+?? 2
+(1-?? )?? +(1+?? )?? -2???? -3?? +2=0
 
Page 3


Edurev123 
4. Sphere and its Properties 
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =?? 
and passing through the ci zle 
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =?? 
(2009 : 12 Marks) 
Solution: 
Approach: General equation of sphere through any circle is used. The parameter san be 
found by the Centre satisfying equation of plane. 
General equation of a sphere passing through the circle is 
?? +???? =0 
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0 
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0 
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if 
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3 
?                                                                                                        3?? +37=6 
?                                                                                                                     ?? =
-31
3
 
? Required sphere is 
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0 
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a 
circle of radius 1 and find the equation of the sphere which has this circle as great 
circle. 
(2010: 12 Marks) 
Solution: 
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3 
 Centre of given circle  =(
1
2
,-
1
2
,0)
 Radius  =
v
1
4
,
1
4
,2=
v
2+
1
2
 =
v
5
2
 
Let ?? be the centre of this circle. 
Distance of plane from centre 
|
1
2
-
1
2
-3|
v1
2
+1
2
+2
2
=
3
v6
 
? Radius of circle with ???? as radius 
v
5
2
-
9
6
=
v
2
2
=1 
 
 
Equation of sphere with circle as great circle. 
                        ?? 2
+?? 2
+?? 2
-?? +?? -2+?? (?? +?? -2?? -3)=0
?                     ?? 2
+?? 2
+?? 2
-?? +?? -2+???? +???? -2???? -?? 3=0
?              ?? 2
+?? 2
+?? 2
+(1-?? )?? +(1+?? )?? -2???? -3?? +2=0
 
Radius of the sphere =1 
?                                          (
1-?? 2
)
2
+(
1+?? 2
)
2
+?? 2
+3?? +2=1
2
?                            
1+?? 2
+2?? 4
+
1+?? 2
-2?? 4
+?? 2
+3?? +2=1
?     
3?? 2
2
+
1
2
+3?? =-1
? 
3?? 2
2
+3?? +
3
2
=0
?   ?? 2
+2?? +1=0
?         (?? +1)
2
=0
?                          ?? =-1
 
Using this value of ?? , the equation of sphere is 
                                     ?? 2
+?? 2
+?? 2
-?? +?? -2-1(?? +?? -2?? -3)=0
?                                           ?? 2
+?? 2
+?? 2
-?? +?? -?? -?? -?? +2?? +3=0
?                                                                        ?? 2
+?? 2
+?? 2
-3?? +2?? +1=0
 
4.3 Show that the equation of the sphere which touches the sphere 
?? (?? ?? +?? ?? +?? ?? )+???? ?? -???? ?? -?? ?? =?? 
at the point (?? ,?? ,-?? ) and passes through the point (-?? ,?? ,?? ) is 
?? ?? +?? ?? +?? ?? +?? ?? -?? ?? +?? =?? 
(2011 : 10 Marks) 
Solution: 
The equation of the given sphere is 
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? =0 (??) 
The equation of the tangent plane to the sphere (i) at the point (1,2,-2) is 
4(?? ·1+?? ·2+?? (-2))+5·(?? +1)-
25
2
(?? +2)-(?? -2)=0
 
Or                                                         18?? -9?? -18?? +14=0                                                           (???? ) 
? The equation of the sphere which touches the sphere (i) at (1,2,-2) is 
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? +?? (18?? -9?? -18?? +14)=0                                        (?????? ) 
If (iii) passes through (-1,0,1) then, 
4(1+0+0)+10(-1)-0-0+?? (-18-0-0+14)=0 
Page 4


Edurev123 
4. Sphere and its Properties 
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =?? 
and passing through the ci zle 
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =?? 
(2009 : 12 Marks) 
Solution: 
Approach: General equation of sphere through any circle is used. The parameter san be 
found by the Centre satisfying equation of plane. 
General equation of a sphere passing through the circle is 
?? +???? =0 
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0 
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0 
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if 
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3 
?                                                                                                        3?? +37=6 
?                                                                                                                     ?? =
-31
3
 
? Required sphere is 
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0 
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a 
circle of radius 1 and find the equation of the sphere which has this circle as great 
circle. 
(2010: 12 Marks) 
Solution: 
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3 
 Centre of given circle  =(
1
2
,-
1
2
,0)
 Radius  =
v
1
4
,
1
4
,2=
v
2+
1
2
 =
v
5
2
 
Let ?? be the centre of this circle. 
Distance of plane from centre 
|
1
2
-
1
2
-3|
v1
2
+1
2
+2
2
=
3
v6
 
? Radius of circle with ???? as radius 
v
5
2
-
9
6
=
v
2
2
=1 
 
 
Equation of sphere with circle as great circle. 
                        ?? 2
+?? 2
+?? 2
-?? +?? -2+?? (?? +?? -2?? -3)=0
?                     ?? 2
+?? 2
+?? 2
-?? +?? -2+???? +???? -2???? -?? 3=0
?              ?? 2
+?? 2
+?? 2
+(1-?? )?? +(1+?? )?? -2???? -3?? +2=0
 
Radius of the sphere =1 
?                                          (
1-?? 2
)
2
+(
1+?? 2
)
2
+?? 2
+3?? +2=1
2
?                            
1+?? 2
+2?? 4
+
1+?? 2
-2?? 4
+?? 2
+3?? +2=1
?     
3?? 2
2
+
1
2
+3?? =-1
? 
3?? 2
2
+3?? +
3
2
=0
?   ?? 2
+2?? +1=0
?         (?? +1)
2
=0
?                          ?? =-1
 
Using this value of ?? , the equation of sphere is 
                                     ?? 2
+?? 2
+?? 2
-?? +?? -2-1(?? +?? -2?? -3)=0
?                                           ?? 2
+?? 2
+?? 2
-?? +?? -?? -?? -?? +2?? +3=0
?                                                                        ?? 2
+?? 2
+?? 2
-3?? +2?? +1=0
 
4.3 Show that the equation of the sphere which touches the sphere 
?? (?? ?? +?? ?? +?? ?? )+???? ?? -???? ?? -?? ?? =?? 
at the point (?? ,?? ,-?? ) and passes through the point (-?? ,?? ,?? ) is 
?? ?? +?? ?? +?? ?? +?? ?? -?? ?? +?? =?? 
(2011 : 10 Marks) 
Solution: 
The equation of the given sphere is 
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? =0 (??) 
The equation of the tangent plane to the sphere (i) at the point (1,2,-2) is 
4(?? ·1+?? ·2+?? (-2))+5·(?? +1)-
25
2
(?? +2)-(?? -2)=0
 
Or                                                         18?? -9?? -18?? +14=0                                                           (???? ) 
? The equation of the sphere which touches the sphere (i) at (1,2,-2) is 
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? +?? (18?? -9?? -18?? +14)=0                                        (?????? ) 
If (iii) passes through (-1,0,1) then, 
4(1+0+0)+10(-1)-0-0+?? (-18-0-0+14)=0 
4-10+?? (-4) =0
-6 =4?? ??? =-
3
2
 
Put ?? =-
3
2
 in (iii), 
                         4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? -
3
2
(18?? -9?? -18?? +14)=0
 
?                        8(?? 2
+?? 2
+?? 2
)+20?? -50?? -4?? -54?? +27?? +54?? -42=0 
?8(?? 2
+?? 2
+?? 2
)-34?? -23?? +50?? -42=0, which is the required equation of the 
sphere. 
4.4 Show that three mutually perpendicular tangent lines can be drawn to the 
sphere ?? ?? +?? ?? +?? ?? =?? ?? from any point on the sphere ?? (?? ?? +?? ?? +?? ?? )=?? ?? ?? . 
(2013 : 15 Marks) 
Solution: 
Let (?? ,?? ,?? ) be any point. Equation of enveloping cone from this point to sphere 
?? 2
+?? 2
+?? 2
=1
2
(??) 
is                                                                                      ?? ?? 1
=?? 1
2
 
?(?? 2
+?? 2
+?? 2
-?? 2
)(?? 2
+?? 2
+?? 2
-?? 2
)=(???? +???? +???? -?? 2
)
2
 
This cone will have three mutually perpendicular generators if coefficient of ?? 2
+ 
coefficient of ?? 2
+ coefficient of ?? 2
=0. 
i.e.,                                                                                        ?? +?? +?? =0 
?           (?? 2
+?? 2
-?? 2
)+(?? 2
+?? 2
-?? 2
)+(?? 2
+?? 2
-?? 2
)=0 
?                                                                             2(?? 2
+?? 2
+?? 2
)=0 
Since this is also the condition that three tangent lines from (?? ,?? ,?? ) to sphere are 
mutually perpendicular, so locus of (?? ,?? ,?? ) is 
2(?? 2
+?? 2
+?? 2
)=3?? 2
 
4.5 Find the co-ordinates of the points on the sphere ?? ?? +?? ?? +?? ?? -?? ?? +?? ?? =?? , 
the tangent planes at which are paraliel to the plane ?? ?? -?? +?? ?? =?? . 
(2014 : 10 Marks) 
Solution: 
Let, the equation of planes ?? 1
 and ?? 2
 parallel to 
Page 5


Edurev123 
4. Sphere and its Properties 
4.1 Find the equation of the sphere having its Centre on the plane ?? ?? -?? ?? -?? =?? 
and passing through the ci zle 
?? ?? +?? ?? +?? ?? -???? ?? -?? ?? +?? ?? +?? =?? ?? ?? +?? ?? -?? ?? +?? =?? 
(2009 : 12 Marks) 
Solution: 
Approach: General equation of sphere through any circle is used. The parameter san be 
found by the Centre satisfying equation of plane. 
General equation of a sphere passing through the circle is 
?? +???? =0 
or (?? 2
+?? 2
+?? 2
-12?? -3?? +4?? +8)+?? (3?? +4?? -5?? +3)=0 
i.e., ?? 2
+?? 2
+?? 2
+(3?? -12)?? +(4?? -3)?? +(4-5?? )?? +3?? +8=0 
The centre of the sphere is (
3?? -12
2
),(
4?? -3
2
),(
4-5?? 2
) . This lies on the given plane if 
-[4(
3?? -12
2
)-5(
4?? -3
2
)-(
4-5?? 2
)] =3 
?                                                                                                        3?? +37=6 
?                                                                                                                     ?? =
-31
3
 
? Required sphere is 
?? 2
+?? 2
+?? 2
-43?? -
133
3
?? +
167
3
?? -23=0 
4.2 Show that the plane ?? +?? -?? ?? =?? cuts the sphere ?? ?? +?? ?? +?? ?? -?? +?? =?? in a 
circle of radius 1 and find the equation of the sphere which has this circle as great 
circle. 
(2010: 12 Marks) 
Solution: 
Given : Equation of circle is ?? 2
+?? 2
+?? 2
-?? +?? =2, Plane =?? +?? -2?? =3 
 Centre of given circle  =(
1
2
,-
1
2
,0)
 Radius  =
v
1
4
,
1
4
,2=
v
2+
1
2
 =
v
5
2
 
Let ?? be the centre of this circle. 
Distance of plane from centre 
|
1
2
-
1
2
-3|
v1
2
+1
2
+2
2
=
3
v6
 
? Radius of circle with ???? as radius 
v
5
2
-
9
6
=
v
2
2
=1 
 
 
Equation of sphere with circle as great circle. 
                        ?? 2
+?? 2
+?? 2
-?? +?? -2+?? (?? +?? -2?? -3)=0
?                     ?? 2
+?? 2
+?? 2
-?? +?? -2+???? +???? -2???? -?? 3=0
?              ?? 2
+?? 2
+?? 2
+(1-?? )?? +(1+?? )?? -2???? -3?? +2=0
 
Radius of the sphere =1 
?                                          (
1-?? 2
)
2
+(
1+?? 2
)
2
+?? 2
+3?? +2=1
2
?                            
1+?? 2
+2?? 4
+
1+?? 2
-2?? 4
+?? 2
+3?? +2=1
?     
3?? 2
2
+
1
2
+3?? =-1
? 
3?? 2
2
+3?? +
3
2
=0
?   ?? 2
+2?? +1=0
?         (?? +1)
2
=0
?                          ?? =-1
 
Using this value of ?? , the equation of sphere is 
                                     ?? 2
+?? 2
+?? 2
-?? +?? -2-1(?? +?? -2?? -3)=0
?                                           ?? 2
+?? 2
+?? 2
-?? +?? -?? -?? -?? +2?? +3=0
?                                                                        ?? 2
+?? 2
+?? 2
-3?? +2?? +1=0
 
4.3 Show that the equation of the sphere which touches the sphere 
?? (?? ?? +?? ?? +?? ?? )+???? ?? -???? ?? -?? ?? =?? 
at the point (?? ,?? ,-?? ) and passes through the point (-?? ,?? ,?? ) is 
?? ?? +?? ?? +?? ?? +?? ?? -?? ?? +?? =?? 
(2011 : 10 Marks) 
Solution: 
The equation of the given sphere is 
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? =0 (??) 
The equation of the tangent plane to the sphere (i) at the point (1,2,-2) is 
4(?? ·1+?? ·2+?? (-2))+5·(?? +1)-
25
2
(?? +2)-(?? -2)=0
 
Or                                                         18?? -9?? -18?? +14=0                                                           (???? ) 
? The equation of the sphere which touches the sphere (i) at (1,2,-2) is 
4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? +?? (18?? -9?? -18?? +14)=0                                        (?????? ) 
If (iii) passes through (-1,0,1) then, 
4(1+0+0)+10(-1)-0-0+?? (-18-0-0+14)=0 
4-10+?? (-4) =0
-6 =4?? ??? =-
3
2
 
Put ?? =-
3
2
 in (iii), 
                         4(?? 2
+?? 2
+?? 2
)+10?? -25?? -2?? -
3
2
(18?? -9?? -18?? +14)=0
 
?                        8(?? 2
+?? 2
+?? 2
)+20?? -50?? -4?? -54?? +27?? +54?? -42=0 
?8(?? 2
+?? 2
+?? 2
)-34?? -23?? +50?? -42=0, which is the required equation of the 
sphere. 
4.4 Show that three mutually perpendicular tangent lines can be drawn to the 
sphere ?? ?? +?? ?? +?? ?? =?? ?? from any point on the sphere ?? (?? ?? +?? ?? +?? ?? )=?? ?? ?? . 
(2013 : 15 Marks) 
Solution: 
Let (?? ,?? ,?? ) be any point. Equation of enveloping cone from this point to sphere 
?? 2
+?? 2
+?? 2
=1
2
(??) 
is                                                                                      ?? ?? 1
=?? 1
2
 
?(?? 2
+?? 2
+?? 2
-?? 2
)(?? 2
+?? 2
+?? 2
-?? 2
)=(???? +???? +???? -?? 2
)
2
 
This cone will have three mutually perpendicular generators if coefficient of ?? 2
+ 
coefficient of ?? 2
+ coefficient of ?? 2
=0. 
i.e.,                                                                                        ?? +?? +?? =0 
?           (?? 2
+?? 2
-?? 2
)+(?? 2
+?? 2
-?? 2
)+(?? 2
+?? 2
-?? 2
)=0 
?                                                                             2(?? 2
+?? 2
+?? 2
)=0 
Since this is also the condition that three tangent lines from (?? ,?? ,?? ) to sphere are 
mutually perpendicular, so locus of (?? ,?? ,?? ) is 
2(?? 2
+?? 2
+?? 2
)=3?? 2
 
4.5 Find the co-ordinates of the points on the sphere ?? ?? +?? ?? +?? ?? -?? ?? +?? ?? =?? , 
the tangent planes at which are paraliel to the plane ?? ?? -?? +?? ?? =?? . 
(2014 : 10 Marks) 
Solution: 
Let, the equation of planes ?? 1
 and ?? 2
 parallel to 
2?? -?? +2?? =1
2?? -?? +2?? +?? =0
2?? -?? +2?? +?? =0
 
Now, of 
be tangent to sphere length perpendicular to ?? 1
 and ?? 2
= radius of sphere. 
|
2(2)-1(-1)+2(0)+?? v4+1+4
|=3 
So, 
?? =14,-4 
                                       
Now, to find points of constant of tangent plane, ?? 1
,?? 2
 and sphere (point ?? and ?? in 
diagram) equation of line ' ?? 1
 ' normal to tangent plane and passing through centre 
(2,-1,0) is 
 
?? -2
2
=
?? +1
-1
=
?? -0
2
=?? 
For ?? and ?? ??? =±?? 
???? ,                                     
?? -2
2/3
=
?? +1
-1/3
=
?? 2/3
=±3 
or                                                                (?? ,?? ,?? )˜(4,-2,2),(0,0,-2) 
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FAQs on Sphere and its Properties - Mathematics Optional Notes for UPSC

1. What are the properties of a sphere?
Ans. A sphere is a three-dimensional geometric shape that is perfectly round in shape. It has properties such as a constant diameter, a constant radius from the center to any point on the surface, and all points on the surface equidistant from the center.
2. How is the volume of a sphere calculated?
Ans. The volume of a sphere can be calculated using the formula V = (4/3)πr^3, where "r" represents the radius of the sphere and π is a constant approximately equal to 3.14159.
3. What is the surface area of a sphere formula?
Ans. The surface area of a sphere can be calculated using the formula A = 4πr^2, where "r" represents the radius of the sphere and π is a constant approximately equal to 3.14159.
4. How are spheres used in real life applications?
Ans. Spheres are used in various real-life applications such as in sports equipment like soccer balls, basketballs, and baseballs. They are also used in architectural design for creating domes and in packaging for designing round containers.
5. Can a sphere have different shapes or dimensions?
Ans. No, a sphere is a specific geometric shape that is perfectly round and symmetrical. It cannot have different shapes or dimensions as its properties are defined by its constant radius and equidistant points from the center.
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Important questions

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Previous Year Questions with Solutions

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Free

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Sphere and its Properties | Mathematics Optional Notes for UPSC

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Sample Paper

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