All Courses
Get the App Signup / Login
Sign in Open App
JEE Exam  >  JEE Notes  >  Mathematics (Maths) for JEE Main & Advanced  >  Solved Examples: Circle

Solved Examples: Circle | Mathematics (Maths) for JEE Main & Advanced PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


Solved Examples on Circles 
Q1: The lines ?? ?? - ?? ?? = ?? and ?? ?? - ?? ?? = ?? are the diameters of a circle of 
area 154 square units. The equation of the circle is 
(a) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(b) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
(c) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(d) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
Ans: (b) 
Sol:  
Centre of circle = Point of intersection of diameters, 
On solving equations, 2?? - 3?? = 5 and 3?? - 4?? = 7, we get, (?? , ?? ) = (1, -1) 
? Centre of circle = (1, -1). Now area of circle = 154 ? ?? ?? 2
= 154 ? ?? = 7 
Hence, the equation of required circle is (?? - 1)
2
+ (?? + 1)
2
= (7)
2
? ?? 2
+ ?? 2
-
2?? + 2?? = 47. 
Q2: A circle of radius 5 units touches both the axes and lies in first 
quadrant. If the circle makes one complete roll on ?? -axis along the positive 
direction of ?? -axis, then its equation in the new position is 
(a) ?? ?? + ?? ?? + ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ???? ???? + ???? ?? + ?????? ?? ?? = ?? 
(c) ?? ?? + ?? ?? - ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(d) None of these 
Ans: (d) 
Sol: 
The ?? -coordinate of the new position of the circle is 5 + circumferrence of the first 
circle = 5 + 10?? The ?? -coordinate is 5 and the radius is also 5 . 
Hence, the equation of the circle in the new position is (?? - 5 - 10?? )
2
+ (?? -
5)
2
= (5)
2
 
 ? ?? 2
+ 25 + 100?? 2
- 10?? + 100?? - 20???? + ?? 2
+ 25 - 10?? = 25
 ? ?? 2
+ ?? 2
- 20???? - 10?? - 10?? + 100?? 2
+ 100?? + 25 = 0
 
Page 2


Solved Examples on Circles 
Q1: The lines ?? ?? - ?? ?? = ?? and ?? ?? - ?? ?? = ?? are the diameters of a circle of 
area 154 square units. The equation of the circle is 
(a) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(b) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
(c) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(d) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
Ans: (b) 
Sol:  
Centre of circle = Point of intersection of diameters, 
On solving equations, 2?? - 3?? = 5 and 3?? - 4?? = 7, we get, (?? , ?? ) = (1, -1) 
? Centre of circle = (1, -1). Now area of circle = 154 ? ?? ?? 2
= 154 ? ?? = 7 
Hence, the equation of required circle is (?? - 1)
2
+ (?? + 1)
2
= (7)
2
? ?? 2
+ ?? 2
-
2?? + 2?? = 47. 
Q2: A circle of radius 5 units touches both the axes and lies in first 
quadrant. If the circle makes one complete roll on ?? -axis along the positive 
direction of ?? -axis, then its equation in the new position is 
(a) ?? ?? + ?? ?? + ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ???? ???? + ???? ?? + ?????? ?? ?? = ?? 
(c) ?? ?? + ?? ?? - ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(d) None of these 
Ans: (d) 
Sol: 
The ?? -coordinate of the new position of the circle is 5 + circumferrence of the first 
circle = 5 + 10?? The ?? -coordinate is 5 and the radius is also 5 . 
Hence, the equation of the circle in the new position is (?? - 5 - 10?? )
2
+ (?? -
5)
2
= (5)
2
 
 ? ?? 2
+ 25 + 100?? 2
- 10?? + 100?? - 20???? + ?? 2
+ 25 - 10?? = 25
 ? ?? 2
+ ?? 2
- 20???? - 10?? - 10?? + 100?? 2
+ 100?? + 25 = 0
 
Q3: The abscissae of ?? and ?? are the roots of the equation ?? ?? + ?? ???? - ?? ?? =
?? and their ordinates are the roots of the equation ?? ?? + ?? ???? - ?? ?? = ?? . The 
equation of the circle with ???? as diameter is 
(a) ?? ?? + ?? ?? + ?? ???? + ?? ???? - ?? ?? - ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ?? ???? + ???? - ?? ?? - ?? ?? = ?? 
(c) ?? ?? + ?? ?? + ?? ???? + ?? ???? + ?? ?? + ?? ?? = ?? 
(d) None of these 
Ans: (a) 
Sol:  
Let ?? 1
, ?? 2
 and ?? 1
, ?? 2
 be roots of ?? 2
+ 2???? - ?? 2
= 0 and ?? 2
+ 2???? - ?? 2
= 0 
respectively. 
Then, ?? 1
+ ?? 2
= -2?? , ?? 1
?? 2
= -?? 2
 and ?? 1
+ ?? 2
= -2?? , ?? 1
?? 2
= -?? 2
 
The equation of the circle with ?? (?? 1
, ?? 1
) and ?? (?? 2
, ?? 2
) as the end points of 
diameter is 
(?? - ?? 1
)(?? - ?? 2
) + (?? - ?? 1
)(?? - ?? 2
) = 0
?? 2
+ ?? 2
- ?? (?? 1
+ ?? 2
) - ?? (?? 1
+ ?? 2
) + ?? 1
?? 2
+ ?? 1
?? 2
= 0; ?? 2
+ ?? 2
+ 2???? + 2???? - ?? 2
- ?? 2
= 0
 
 
Q4: If the straight line ?? = ???? is outside the circle ?? ?? + ?? ?? - ???? ?? + ???? = ?? , 
then 
(a) ?? > ?? 
(b) ?? < ?? 
(c) |?? | > ?? 
(d) |?? | < ?? 
Ans: (d) 
Sol: If the straight line ?? = ???? is outside the given circle then perpendicular 
distance of line from centre of circle > radius of circle 
10
v1 + ?? 2
> v10 ? (1 + ?? 2
) < 10 ? ?? 2
< 9 ? |?? | < 3 
Page 3


Solved Examples on Circles 
Q1: The lines ?? ?? - ?? ?? = ?? and ?? ?? - ?? ?? = ?? are the diameters of a circle of 
area 154 square units. The equation of the circle is 
(a) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(b) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
(c) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(d) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
Ans: (b) 
Sol:  
Centre of circle = Point of intersection of diameters, 
On solving equations, 2?? - 3?? = 5 and 3?? - 4?? = 7, we get, (?? , ?? ) = (1, -1) 
? Centre of circle = (1, -1). Now area of circle = 154 ? ?? ?? 2
= 154 ? ?? = 7 
Hence, the equation of required circle is (?? - 1)
2
+ (?? + 1)
2
= (7)
2
? ?? 2
+ ?? 2
-
2?? + 2?? = 47. 
Q2: A circle of radius 5 units touches both the axes and lies in first 
quadrant. If the circle makes one complete roll on ?? -axis along the positive 
direction of ?? -axis, then its equation in the new position is 
(a) ?? ?? + ?? ?? + ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ???? ???? + ???? ?? + ?????? ?? ?? = ?? 
(c) ?? ?? + ?? ?? - ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(d) None of these 
Ans: (d) 
Sol: 
The ?? -coordinate of the new position of the circle is 5 + circumferrence of the first 
circle = 5 + 10?? The ?? -coordinate is 5 and the radius is also 5 . 
Hence, the equation of the circle in the new position is (?? - 5 - 10?? )
2
+ (?? -
5)
2
= (5)
2
 
 ? ?? 2
+ 25 + 100?? 2
- 10?? + 100?? - 20???? + ?? 2
+ 25 - 10?? = 25
 ? ?? 2
+ ?? 2
- 20???? - 10?? - 10?? + 100?? 2
+ 100?? + 25 = 0
 
Q3: The abscissae of ?? and ?? are the roots of the equation ?? ?? + ?? ???? - ?? ?? =
?? and their ordinates are the roots of the equation ?? ?? + ?? ???? - ?? ?? = ?? . The 
equation of the circle with ???? as diameter is 
(a) ?? ?? + ?? ?? + ?? ???? + ?? ???? - ?? ?? - ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ?? ???? + ???? - ?? ?? - ?? ?? = ?? 
(c) ?? ?? + ?? ?? + ?? ???? + ?? ???? + ?? ?? + ?? ?? = ?? 
(d) None of these 
Ans: (a) 
Sol:  
Let ?? 1
, ?? 2
 and ?? 1
, ?? 2
 be roots of ?? 2
+ 2???? - ?? 2
= 0 and ?? 2
+ 2???? - ?? 2
= 0 
respectively. 
Then, ?? 1
+ ?? 2
= -2?? , ?? 1
?? 2
= -?? 2
 and ?? 1
+ ?? 2
= -2?? , ?? 1
?? 2
= -?? 2
 
The equation of the circle with ?? (?? 1
, ?? 1
) and ?? (?? 2
, ?? 2
) as the end points of 
diameter is 
(?? - ?? 1
)(?? - ?? 2
) + (?? - ?? 1
)(?? - ?? 2
) = 0
?? 2
+ ?? 2
- ?? (?? 1
+ ?? 2
) - ?? (?? 1
+ ?? 2
) + ?? 1
?? 2
+ ?? 1
?? 2
= 0; ?? 2
+ ?? 2
+ 2???? + 2???? - ?? 2
- ?? 2
= 0
 
 
Q4: If the straight line ?? = ???? is outside the circle ?? ?? + ?? ?? - ???? ?? + ???? = ?? , 
then 
(a) ?? > ?? 
(b) ?? < ?? 
(c) |?? | > ?? 
(d) |?? | < ?? 
Ans: (d) 
Sol: If the straight line ?? = ???? is outside the given circle then perpendicular 
distance of line from centre of circle > radius of circle 
10
v1 + ?? 2
> v10 ? (1 + ?? 2
) < 10 ? ?? 2
< 9 ? |?? | < 3 
Q5: The equations to the tangents to the circle ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? which 
are parallel to the straight line ???? + ???? + ?? = ?? , are 
(a) ?? ?? - ?? ?? - ???? = ?? , ?? ?? - ?? ?? + ???? = ?? 
(b) ?? ?? + ?? ?? - ???? = ?? , ?? ?? + ?? ?? + ???? = ?? 
(c) ?? ?? + ?? ?? + ???? = ?? , ?? ?? + ?? ?? - ???? = ?? 
(d) ?? ?? - ?? ?? + ???? = ?? , ?? ?? - ?? ?? + ???? = ?? 
Ans: (c) 
Sol: Let equation of tangent be 4?? + 3?? + ?? = 0,  
then v 9 + 4 + 12 = |
4(3)+3(-2)+?? v 16+9
| ? 6 + ?? = ±25 ? ?? = 19 and -31 
Hence the equations of tangents are 4?? + 3?? + 19 = 0 and 4?? + 3?? - 31 = 0 
Q6: If the distances from the origin to the centres of three circles ?? ?? + ?? ?? +
?? ?? ?? ?? - ?? ?? = ?? (?? = ?? , ?? , ?? ) are in G.P. then the lengths of the tangents drawn 
to them from any point on the circle ?? ?? + ?? ?? = ?? ?? are in 
(a) A.P. 
(b) G.P. 
(c) H.P. 
(d) None of these 
Ans: (b) 
Sol: The centres of the given circles are (-?? ?? , 0)(?? = 1,2,3) 
The distances from the origin to the centres are ?? ?? (?? = 1,2,3). It is given that ?? 2
2
=
?? 1
?? 3
. 
Let ?? (h, ?? ) be any point on the circle ?? 2
+ ?? 2
= ?? 2
, then, h
2
+ ?? 2
= ?? 2
Now, ?? ?? = 
length of the tangent from (h, ?? ) to ?? 2
+ ?? 2
+ 2?? ?? ?? - ?? 2
= 0 
= vh
2
+ ?? 2
+ 2?? ?? h - ?? 2
= v?? 2
+ 2?? ?? h - ?? 2
= v2?? ?? h [? h
2
+ ?? 2
= ?? 2
 and ?? = 1,2,3] 
Therefore, ?? 2
2
= 2?? 2
h = 2h(v?? 1
?? 3
)       
[? ?? 2
2
= ?? 1
?? 3
]     
= v2h?? 1
v2h?? 3
= ?? 1
?? 3
.  
Hence, ?? 1
, ?? 2
, ?? 3
 are in G.P. 
Page 4


Solved Examples on Circles 
Q1: The lines ?? ?? - ?? ?? = ?? and ?? ?? - ?? ?? = ?? are the diameters of a circle of 
area 154 square units. The equation of the circle is 
(a) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(b) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
(c) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(d) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
Ans: (b) 
Sol:  
Centre of circle = Point of intersection of diameters, 
On solving equations, 2?? - 3?? = 5 and 3?? - 4?? = 7, we get, (?? , ?? ) = (1, -1) 
? Centre of circle = (1, -1). Now area of circle = 154 ? ?? ?? 2
= 154 ? ?? = 7 
Hence, the equation of required circle is (?? - 1)
2
+ (?? + 1)
2
= (7)
2
? ?? 2
+ ?? 2
-
2?? + 2?? = 47. 
Q2: A circle of radius 5 units touches both the axes and lies in first 
quadrant. If the circle makes one complete roll on ?? -axis along the positive 
direction of ?? -axis, then its equation in the new position is 
(a) ?? ?? + ?? ?? + ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ???? ???? + ???? ?? + ?????? ?? ?? = ?? 
(c) ?? ?? + ?? ?? - ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(d) None of these 
Ans: (d) 
Sol: 
The ?? -coordinate of the new position of the circle is 5 + circumferrence of the first 
circle = 5 + 10?? The ?? -coordinate is 5 and the radius is also 5 . 
Hence, the equation of the circle in the new position is (?? - 5 - 10?? )
2
+ (?? -
5)
2
= (5)
2
 
 ? ?? 2
+ 25 + 100?? 2
- 10?? + 100?? - 20???? + ?? 2
+ 25 - 10?? = 25
 ? ?? 2
+ ?? 2
- 20???? - 10?? - 10?? + 100?? 2
+ 100?? + 25 = 0
 
Q3: The abscissae of ?? and ?? are the roots of the equation ?? ?? + ?? ???? - ?? ?? =
?? and their ordinates are the roots of the equation ?? ?? + ?? ???? - ?? ?? = ?? . The 
equation of the circle with ???? as diameter is 
(a) ?? ?? + ?? ?? + ?? ???? + ?? ???? - ?? ?? - ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ?? ???? + ???? - ?? ?? - ?? ?? = ?? 
(c) ?? ?? + ?? ?? + ?? ???? + ?? ???? + ?? ?? + ?? ?? = ?? 
(d) None of these 
Ans: (a) 
Sol:  
Let ?? 1
, ?? 2
 and ?? 1
, ?? 2
 be roots of ?? 2
+ 2???? - ?? 2
= 0 and ?? 2
+ 2???? - ?? 2
= 0 
respectively. 
Then, ?? 1
+ ?? 2
= -2?? , ?? 1
?? 2
= -?? 2
 and ?? 1
+ ?? 2
= -2?? , ?? 1
?? 2
= -?? 2
 
The equation of the circle with ?? (?? 1
, ?? 1
) and ?? (?? 2
, ?? 2
) as the end points of 
diameter is 
(?? - ?? 1
)(?? - ?? 2
) + (?? - ?? 1
)(?? - ?? 2
) = 0
?? 2
+ ?? 2
- ?? (?? 1
+ ?? 2
) - ?? (?? 1
+ ?? 2
) + ?? 1
?? 2
+ ?? 1
?? 2
= 0; ?? 2
+ ?? 2
+ 2???? + 2???? - ?? 2
- ?? 2
= 0
 
 
Q4: If the straight line ?? = ???? is outside the circle ?? ?? + ?? ?? - ???? ?? + ???? = ?? , 
then 
(a) ?? > ?? 
(b) ?? < ?? 
(c) |?? | > ?? 
(d) |?? | < ?? 
Ans: (d) 
Sol: If the straight line ?? = ???? is outside the given circle then perpendicular 
distance of line from centre of circle > radius of circle 
10
v1 + ?? 2
> v10 ? (1 + ?? 2
) < 10 ? ?? 2
< 9 ? |?? | < 3 
Q5: The equations to the tangents to the circle ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? which 
are parallel to the straight line ???? + ???? + ?? = ?? , are 
(a) ?? ?? - ?? ?? - ???? = ?? , ?? ?? - ?? ?? + ???? = ?? 
(b) ?? ?? + ?? ?? - ???? = ?? , ?? ?? + ?? ?? + ???? = ?? 
(c) ?? ?? + ?? ?? + ???? = ?? , ?? ?? + ?? ?? - ???? = ?? 
(d) ?? ?? - ?? ?? + ???? = ?? , ?? ?? - ?? ?? + ???? = ?? 
Ans: (c) 
Sol: Let equation of tangent be 4?? + 3?? + ?? = 0,  
then v 9 + 4 + 12 = |
4(3)+3(-2)+?? v 16+9
| ? 6 + ?? = ±25 ? ?? = 19 and -31 
Hence the equations of tangents are 4?? + 3?? + 19 = 0 and 4?? + 3?? - 31 = 0 
Q6: If the distances from the origin to the centres of three circles ?? ?? + ?? ?? +
?? ?? ?? ?? - ?? ?? = ?? (?? = ?? , ?? , ?? ) are in G.P. then the lengths of the tangents drawn 
to them from any point on the circle ?? ?? + ?? ?? = ?? ?? are in 
(a) A.P. 
(b) G.P. 
(c) H.P. 
(d) None of these 
Ans: (b) 
Sol: The centres of the given circles are (-?? ?? , 0)(?? = 1,2,3) 
The distances from the origin to the centres are ?? ?? (?? = 1,2,3). It is given that ?? 2
2
=
?? 1
?? 3
. 
Let ?? (h, ?? ) be any point on the circle ?? 2
+ ?? 2
= ?? 2
, then, h
2
+ ?? 2
= ?? 2
Now, ?? ?? = 
length of the tangent from (h, ?? ) to ?? 2
+ ?? 2
+ 2?? ?? ?? - ?? 2
= 0 
= vh
2
+ ?? 2
+ 2?? ?? h - ?? 2
= v?? 2
+ 2?? ?? h - ?? 2
= v2?? ?? h [? h
2
+ ?? 2
= ?? 2
 and ?? = 1,2,3] 
Therefore, ?? 2
2
= 2?? 2
h = 2h(v?? 1
?? 3
)       
[? ?? 2
2
= ?? 1
?? 3
]     
= v2h?? 1
v2h?? 3
= ?? 1
?? 3
.  
Hence, ?? 1
, ?? 2
, ?? 3
 are in G.P. 
Q7: Two tangents to the circle ?? ?? + ?? ?? = ?? at the points ?? and ?? meet at 
?? (-?? , ?? ). The area of quadrilateral ???????? , where ?? is the origin, is 
(a) 4 
(b) ?? v ?? 
(c) ?? v ?? 
(d) None of these 
Ans: (c) 
Sol: 
 
 
Clearly, sin ?? =
2
4
=
1
2
, ? ?? = 30
°
. So area (? ?????? ) =
1
2
· 2 · 4 · sin 60
°
 
? Area (quadrilateral ???? ) = 2 ·
1
2
· 2 · 4sin 60
°
= 8 ·
v 3
2
= 4v 3. 
Trick : Area of quadrilateral = ?? v?? 1
= 2v 12 = 4v 3 
Q8: The area of the triangle formed by the tangents from the point (?? , ?? ) to 
the circle ?? ?? + ?? ?? = ?? and the line joining their points of contact is 
(a) 
????
??????
 sq.units 
(b) 
??????
????
 sq.units 
(c) 
??????
????
 sq.units 
(d) None of these 
Ans: (b) 
Sol:   
Page 5


Solved Examples on Circles 
Q1: The lines ?? ?? - ?? ?? = ?? and ?? ?? - ?? ?? = ?? are the diameters of a circle of 
area 154 square units. The equation of the circle is 
(a) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(b) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
(c) ?? ?? + ?? ?? + ?? ?? - ?? ?? = ???? 
(d) ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? 
Ans: (b) 
Sol:  
Centre of circle = Point of intersection of diameters, 
On solving equations, 2?? - 3?? = 5 and 3?? - 4?? = 7, we get, (?? , ?? ) = (1, -1) 
? Centre of circle = (1, -1). Now area of circle = 154 ? ?? ?? 2
= 154 ? ?? = 7 
Hence, the equation of required circle is (?? - 1)
2
+ (?? + 1)
2
= (7)
2
? ?? 2
+ ?? 2
-
2?? + 2?? = 47. 
Q2: A circle of radius 5 units touches both the axes and lies in first 
quadrant. If the circle makes one complete roll on ?? -axis along the positive 
direction of ?? -axis, then its equation in the new position is 
(a) ?? ?? + ?? ?? + ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ???? ???? + ???? ?? + ?????? ?? ?? = ?? 
(c) ?? ?? + ?? ?? - ???? ???? - ???? ?? + ?????? ?? ?? = ?? 
(d) None of these 
Ans: (d) 
Sol: 
The ?? -coordinate of the new position of the circle is 5 + circumferrence of the first 
circle = 5 + 10?? The ?? -coordinate is 5 and the radius is also 5 . 
Hence, the equation of the circle in the new position is (?? - 5 - 10?? )
2
+ (?? -
5)
2
= (5)
2
 
 ? ?? 2
+ 25 + 100?? 2
- 10?? + 100?? - 20???? + ?? 2
+ 25 - 10?? = 25
 ? ?? 2
+ ?? 2
- 20???? - 10?? - 10?? + 100?? 2
+ 100?? + 25 = 0
 
Q3: The abscissae of ?? and ?? are the roots of the equation ?? ?? + ?? ???? - ?? ?? =
?? and their ordinates are the roots of the equation ?? ?? + ?? ???? - ?? ?? = ?? . The 
equation of the circle with ???? as diameter is 
(a) ?? ?? + ?? ?? + ?? ???? + ?? ???? - ?? ?? - ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ?? ???? + ???? - ?? ?? - ?? ?? = ?? 
(c) ?? ?? + ?? ?? + ?? ???? + ?? ???? + ?? ?? + ?? ?? = ?? 
(d) None of these 
Ans: (a) 
Sol:  
Let ?? 1
, ?? 2
 and ?? 1
, ?? 2
 be roots of ?? 2
+ 2???? - ?? 2
= 0 and ?? 2
+ 2???? - ?? 2
= 0 
respectively. 
Then, ?? 1
+ ?? 2
= -2?? , ?? 1
?? 2
= -?? 2
 and ?? 1
+ ?? 2
= -2?? , ?? 1
?? 2
= -?? 2
 
The equation of the circle with ?? (?? 1
, ?? 1
) and ?? (?? 2
, ?? 2
) as the end points of 
diameter is 
(?? - ?? 1
)(?? - ?? 2
) + (?? - ?? 1
)(?? - ?? 2
) = 0
?? 2
+ ?? 2
- ?? (?? 1
+ ?? 2
) - ?? (?? 1
+ ?? 2
) + ?? 1
?? 2
+ ?? 1
?? 2
= 0; ?? 2
+ ?? 2
+ 2???? + 2???? - ?? 2
- ?? 2
= 0
 
 
Q4: If the straight line ?? = ???? is outside the circle ?? ?? + ?? ?? - ???? ?? + ???? = ?? , 
then 
(a) ?? > ?? 
(b) ?? < ?? 
(c) |?? | > ?? 
(d) |?? | < ?? 
Ans: (d) 
Sol: If the straight line ?? = ???? is outside the given circle then perpendicular 
distance of line from centre of circle > radius of circle 
10
v1 + ?? 2
> v10 ? (1 + ?? 2
) < 10 ? ?? 2
< 9 ? |?? | < 3 
Q5: The equations to the tangents to the circle ?? ?? + ?? ?? - ?? ?? + ?? ?? = ???? which 
are parallel to the straight line ???? + ???? + ?? = ?? , are 
(a) ?? ?? - ?? ?? - ???? = ?? , ?? ?? - ?? ?? + ???? = ?? 
(b) ?? ?? + ?? ?? - ???? = ?? , ?? ?? + ?? ?? + ???? = ?? 
(c) ?? ?? + ?? ?? + ???? = ?? , ?? ?? + ?? ?? - ???? = ?? 
(d) ?? ?? - ?? ?? + ???? = ?? , ?? ?? - ?? ?? + ???? = ?? 
Ans: (c) 
Sol: Let equation of tangent be 4?? + 3?? + ?? = 0,  
then v 9 + 4 + 12 = |
4(3)+3(-2)+?? v 16+9
| ? 6 + ?? = ±25 ? ?? = 19 and -31 
Hence the equations of tangents are 4?? + 3?? + 19 = 0 and 4?? + 3?? - 31 = 0 
Q6: If the distances from the origin to the centres of three circles ?? ?? + ?? ?? +
?? ?? ?? ?? - ?? ?? = ?? (?? = ?? , ?? , ?? ) are in G.P. then the lengths of the tangents drawn 
to them from any point on the circle ?? ?? + ?? ?? = ?? ?? are in 
(a) A.P. 
(b) G.P. 
(c) H.P. 
(d) None of these 
Ans: (b) 
Sol: The centres of the given circles are (-?? ?? , 0)(?? = 1,2,3) 
The distances from the origin to the centres are ?? ?? (?? = 1,2,3). It is given that ?? 2
2
=
?? 1
?? 3
. 
Let ?? (h, ?? ) be any point on the circle ?? 2
+ ?? 2
= ?? 2
, then, h
2
+ ?? 2
= ?? 2
Now, ?? ?? = 
length of the tangent from (h, ?? ) to ?? 2
+ ?? 2
+ 2?? ?? ?? - ?? 2
= 0 
= vh
2
+ ?? 2
+ 2?? ?? h - ?? 2
= v?? 2
+ 2?? ?? h - ?? 2
= v2?? ?? h [? h
2
+ ?? 2
= ?? 2
 and ?? = 1,2,3] 
Therefore, ?? 2
2
= 2?? 2
h = 2h(v?? 1
?? 3
)       
[? ?? 2
2
= ?? 1
?? 3
]     
= v2h?? 1
v2h?? 3
= ?? 1
?? 3
.  
Hence, ?? 1
, ?? 2
, ?? 3
 are in G.P. 
Q7: Two tangents to the circle ?? ?? + ?? ?? = ?? at the points ?? and ?? meet at 
?? (-?? , ?? ). The area of quadrilateral ???????? , where ?? is the origin, is 
(a) 4 
(b) ?? v ?? 
(c) ?? v ?? 
(d) None of these 
Ans: (c) 
Sol: 
 
 
Clearly, sin ?? =
2
4
=
1
2
, ? ?? = 30
°
. So area (? ?????? ) =
1
2
· 2 · 4 · sin 60
°
 
? Area (quadrilateral ???? ) = 2 ·
1
2
· 2 · 4sin 60
°
= 8 ·
v 3
2
= 4v 3. 
Trick : Area of quadrilateral = ?? v?? 1
= 2v 12 = 4v 3 
Q8: The area of the triangle formed by the tangents from the point (?? , ?? ) to 
the circle ?? ?? + ?? ?? = ?? and the line joining their points of contact is 
(a) 
????
??????
 sq.units 
(b) 
??????
????
 sq.units 
(c) 
??????
????
 sq.units 
(d) None of these 
Ans: (b) 
Sol:   
 
 
The equation of the chord of contact of tangents drawn from ?? (4,3) to ?? 2
+ ?? 2
=
9 is 4?? + 3?? = 9. The equation of ???? is ?? =
3
4
?? . 
Now, ???? = (length of the perpendicular from (0,0) on 4?? + 3?? - 9 = 0) =
9
5
 
? ???? = 2. ???? = 2v?? ?? 2
- ?? ?? 2
= 2
v
9 -
81
25
=
24
5
 
Now, ???? = ???? - ???? = 5 -
9
5
=
16
5
.So, Area of ? ?????? =
1
2
(
24
5
) (
16
5
) =
192
25
 sq. units 
Q9: If two distinct chords, drawn from the point (?? , ?? ) on the circle ?? ?? + ?? ?? =
???? + ???? (where , ?? ? ?? ) are bisected by the ?? -axis, then  
(a) ?? ?? = ?? ?? 
(b) ?? ?? = ?? ?? ?? 
(c) ?? ?? < ?? ?? ?? 
(d) ?? ?? > ?? ?? ?? 
Ans: (d) 
Sol: Let (h, 0) be a point on ?? -axis, then the equation of chord whose mid-point is 
(h, 0) will be ?? h -
1
2
?? (?? + h) -
1
2
?? (?? + 0) = h
2
- ?? h. This passes through (?? , ?? ), 
hence ?? h -
1
2
?? (?? + h) -
1
2
?? · ?? = h
2
- ?? h
Read More
209 videos|443 docs|143 tests
Join Course for Free

Top Courses for JEE

FAQs on Solved Examples: Circle - Mathematics (Maths) for JEE Main & Advanced

1. What is the formula to calculate the circumference of a circle?
Ans. The formula to calculate the circumference of a circle is C = 2πr, where C represents the circumference and r is the radius of the circle.
2. How is the area of a circle calculated?
Ans. The area of a circle is calculated using the formula A = πr^2, where A represents the area and r is the radius of the circle.
3. How can the diameter of a circle be determined if the circumference is known?
Ans. The diameter of a circle can be determined by dividing the circumference by π, which is approximately 3.14159. The formula is d = C/π, where d represents the diameter and C is the circumference.
4. What is the relationship between the radius and the diameter of a circle?
Ans. The diameter of a circle is twice the length of the radius. In other words, the diameter is equal to 2 times the radius, represented by the formula d = 2r.
5. How can the radius of a circle be calculated if the diameter is known?
Ans. The radius of a circle can be calculated by dividing the diameter by 2. The formula to find the radius is r = d/2, where r represents the radius and d is the diameter of the circle.
Related Exams
JEE
About this Document
4.60/5 Rating
Dec 18, 2024 Last updated
Document Description: Solved Examples: Circle for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The notes and questions for Solved Examples: Circle have been prepared according to the JEE exam syllabus. Information about Solved Examples: Circle covers topics like and Solved Examples: Circle Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Solved Examples: Circle.
Introduction of Solved Examples: Circle in English is available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Solved Examples: Circle in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: Solved Examples: Circle | Mathematics (Maths) for JEE Main & Advanced
Description
Full syllabus notes, lecture & questions for Solved Examples: Circle | Mathematics (Maths) for JEE Main & Advanced - JEE | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) for JEE Main & Advanced | Best notes, free PDF download
Information about Solved Examples: Circle
In this doc you can find the meaning of Solved Examples: Circle defined & explained in the simplest way possible. Besides explaining types of Solved Examples: Circle theory, EduRev gives you an ample number of questions to practice Solved Examples: Circle tests, examples and also practice JEE tests
209 videos|443 docs|143 tests
Join Course for Free
Download as PDF
Explore Courses for JEE exam

Top Courses for JEE

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Sample Paper

,

ppt

,

Summary

,

Objective type Questions

,

Semester Notes

,

Free

,

video lectures

,

Viva Questions

,

past year papers

,

MCQs

,

Important questions

,

Exam

,

Solved Examples: Circle | Mathematics (Maths) for JEE Main & Advanced

,

Previous Year Questions with Solutions

,

Extra Questions

,

Solved Examples: Circle | Mathematics (Maths) for JEE Main & Advanced

,

practice quizzes

,

shortcuts and tricks

,

Solved Examples: Circle | Mathematics (Maths) for JEE Main & Advanced

,

mock tests for examination

,

study material

,

pdf

;
Additional Information about Solved Examples: Circle for JEE Preparation

Solved Examples: Circle Free PDF Download

The Solved Examples: Circle is an invaluable resource that delves deep into the core of the JEE exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the Solved Examples: Circle now and kickstart your journey towards success in the JEE exam.

Importance of Solved Examples: Circle

The importance of Solved Examples: Circle cannot be overstated, especially for JEE aspirants. This document holds the key to success in the JEE exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

Solved Examples: Circle Notes

Solved Examples: Circle Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to Solved Examples: Circle. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, Solved Examples: Circle Notes on EduRev are your ultimate resource for success.

Solved Examples: Circle JEE Questions

The "Solved Examples: Circle JEE Questions" guide is a valuable resource for all aspiring students preparing for the JEE exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study Solved Examples: Circle on the App

Students of JEE can study Solved Examples: Circle alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the Solved Examples: Circle, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of Solved Examples: Circle is prepared as per the latest JEE syllabus.
© EduRev
Education Revolution
Follow Us
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup