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 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
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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.
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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.
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