1. 1/3
GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
Target : JEE (Main + Advanced) 2021
GR_PT (FOR JEE-ADV.) # 09 (Straight line + Circle) MATHEMATICS
SOLUTION
SECTION–I(i)
1. Ans. (B)
2
3
2 2
2
a + a +
=
2
3 4
a + a + =
2
1 0
a + a - = Þ
1 5
2
+
a = - or
1 5
2
-
-
a has only one value in first quadrant.
2. Ans. (B)
Family of lines can be represented as
a(x + 2y – 5) + b(3x – y – 1) = 0
Hence the family always pass through the point
(1,2).
Equation of other bisector must be of the form
4x + 3y = k. (Q angle bisectors are
perpendicular to each other)
As it pass through the point (1, 2) we get k =
10.
3. Ans. (D)
Area will be maximum when the line x + y + 1
= 0 is the diameter of the given circle.
Hence g
2
+ ƒ
2
= 1
Hence (g, ƒ ) can take the values (0, ±1) or (±1,
0)
4. Ans. (B)
We have, a + 2h + b = 0
Compairing with ax + by = h we get (x, y)
as
1 1
,
2 2
æ ö
- -
ç ÷
è ø
5. Ans. (A)
L1
: (x – y – 6) + l (2x + y + 3)
Always passes through A(1, –5)
and
L2
: (x + 2y – 4) + m(–3x + 2y + 4) = 0
Always passes through B(2, 1)
Since lines intersect at right angle so locus of
point of intersection of L1
and L2
will be a circle
with diametric end points A and B.
Þ x2
+ y2
– 3x + 4y – 3 = 0
6. Ans. (A,B,C,D)
For equilateral D
2 2 2
1 2 3 1 2 2 3 3 1
z z z z z z z z z
+ + = + +
(A) Which is satisfied by –z1– z2 & z3 also
similarly |z1| = |z2| = |z3| = 1
(B) z1 + 1, z2 + 1, z3 + 1 are also satisfying
the
relation
(C)
z1
z2
z3
z +z
2
1 3
z +z
2
1 2
z +z
2
2 3
are also vertices of equilateral D.
(D) 3
1 2 z
z z
, ,
2 2 2
are also vertices of an
equilateral triangle.
7. Ans. (A,D)
For ax + by + c = 0 & c = –2a – 3b
Þ a(x – 2) + b(y – 3) = 0
family always passes through (2, 3)
Let equation of tangent is y =
1
mx
m
+
3 = 2m +
1
m
Þ 2m2
– 3m + 1 = 0
(2m – 1)(m – 1) = 0
1
m
2
= & m = 1
Tangents are y =
x
2
2
+ Þ 2y = x + 4
y = x + 1
8. Ans.(A,B,C,D)
ƒ '(x) = 4x3 + kx2 + 4x + k
= (4x + k) (x2 + 1)
(A) If x Î (0,2) then k < ƒ '(x) < 5 (8 + k)
ƒ '(x) > 0 Þ k > 0
(B) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8
(C) If x Î (–2,2)
5(–8 + k) < ƒ '(x) < 5(8 + k)
ƒ '(x) > 0 Þ 5(–8 + k) > 0
or k > 8
2. GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
2/3
Target : JEE (Main + Advanced) 2021
(D) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8
SECTION–I(iii)
Paragraph for Question 9 to 10
Radius of Ci = 1
Radius of C = 2 1
+
C2 C1
C3
C4
C
9. Ans. (B)
Required probability
( )
2
4.
. 2 1
p
=
p +
( )
4
4 3 2 2
3 2 2
= = -
+
10. Ans. (C)
Let the hyperbola be
2 2
2 2
x y
1
a 2a
- =
Passing through ( ) ( )
( )
2 1 , 2 1
± + ± +
Þ ( )
2
2
2 1 2a
+ =
lengths of L.R =
2
2.2a
a
2 1
4
2
æ ö
+
= ç ÷
ç ÷
è ø
Paragraph for Question 11 to 13
11. Ans.(D)
(–2,0)
A
1
0 (2,0)
S1
B
X
S2
Y
1
(x2
+ y2
+ 3)2
– (4x)2
= 0
Þ S1
: x2
+ y2
– 4x + 3 = 0
S2
: x2
+ y2
+ 4x + 3 = 0
Maximumvalueof
(x1
– x2
)2
+ (y1
– y2
)2
is correspondingto points A
& B
AB = 6
Þ (AB)2
= 36
12. Ans.(B)
q=30°
q=30°
1
(2,0)
(0,0)
1
sin 30
2
q = Þ q = °
Þ Angle = 60°
13. Ans.. (D)
Paragraph for Question 14 to 16
Foot of perpendicular from focus on any of the
tangent always lies on tangent at vertex
h 1 k
,
2 2
+
æ ö
ç ÷
è ø
lies on x = 0
Þ h = – 1
Þ locus is x = –1
P
(h,k) L
S(1,0)
A
(at ,2at)
2
14. Ans. (D)
x = –1
15. Ans. (B)
k
0,
2
æ ö
ç ÷
è ø
also lies on L & tangent at vertex
yt = x + t
2
A º (0, t) Þ k = 2t
(h, k) º (–1, 2t)
y = 3x 4 3 3
+ + is tangent to circle
Þ Perpendicular distance = radius
2
2t 3 4 3 3
1 t
2
+ - -
= +
Þ t 3
= Þ radius = 2
16. Ans. (C)
2
1 t 4
+ = Þ t 15
=
Slope of the tangent Þ
1 1
t 15
=
SECTION–III(i)
1. Ans. 1
1
n 1 2
n
I
0 II
I x .x 1 x dx
-
= -
ò 1
4
24
3
Applying By parts
1 1
n 1 2 3/2 n 2 2 3/2
n
0
0
x (1 x ) (n 1)x (1 x )
I dx
3 3
- -
æ ö
- - -
= - +
ç ÷
è ø
ò
1
n 2 2 2
n
0
(n 1)
I x (1 x ) 1 x dx
3
-
-
= - -
ò
3. 3/3
GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
Target : JEE (Main + Advanced) 2021
n n 2 n
3I (n 1)I (n 1)I
-
= - - - Þ n
n 2
I n 1
I n 2
-
-
=
+
2. Ans. 3
Let eqution of circle be
(x2 + y2 + 2x – 4y – 4) + l(2x – 3y)
Its centre
3
1 ,2
2
l
æ ö
- -l +
ç ÷
è ø
Centre is at minimum distance from (–5,8)
Þ Centre = (–5,8) Þ l = 4
Hence equation of circle is
x2 + y2 + 10x – 16y – 4 = 0
radius 93
=
N
3
31
=
3. Ans. 4
2 5
Ö 90º
B
Ö5
D
A(3,1)
C
ÐABC = 90º
BC 2 5
=
AB 9 1 5 5
= + - =
Þ AC = 5
Power of point Þ AB2
= AD.AC
Þ AD = 1 Þ CD = 4
SECTION–IV
1. Ans. (A)®(R); (B)®(P,Q,R,S); (C)®
(P,Q,S); (D)®(T)
(A) (5!6!.....9!) (10!....14!) (10!...19!)
(20!....24!) (25!.....29!)
In 5!,6!.........,9! exponent of 5 is one.
10!.........14! exponent of 5 is two.
15!.........19! exponent of 5 is 3.
20!.........24! exponent of 5 is 4.
25!.........29! exponent of 5 is 6.
hence it is
5
5
5
10
5
15
5
20
5
30
= 5
80
(B)
( )
2
x
y
1 x
=
+
( )
n n
2
2 2
n 0
|x|dx xdx
2 n 1 n
1 x (1 x )
-
= = +
+ +
ò ò l
A(n) = ln(1 + n
2
) &
2
2
1 1 n
A n
n n
æ ö
+
æ ö
= ç ÷
ç ÷
è ø è ø
l
Þ A(n) –
1
A 2 nn
n
æ ö
= +
ç ÷
è ø
l
Þ
1
A(n) A
n
n 2n
nn
æ ö
- ç ÷
è ø =
l
(even integer)
(C) 2
2
g(x)
(ƒ(x))
= Þ 3
4
g'(x) ƒ'(x)
(ƒ(x))
= -
Þ
2
3 4
ƒ''(x) 3(ƒ'(x))
g''(x) 4
(ƒ(x)) (ƒ(x))
æ ö
= - -
ç ÷
ç ÷
è ø
Þ
3 2
3 4
ƒ''(x) 3(g'(x)(ƒ(x) )
g''(x) 4
(ƒ(x)) 16ƒ(x)
æ ö
= - -
ç ÷
ç ÷
è ø
=
2 2
3
ƒ''(x) 3
4 (g'(x)) (ƒ(x))
16
(ƒ(x))
æ ö
- -
ç ÷
è ø
l = 3
(D) Radical axis of both the circles is x – y = 0
Length of perpendicular from
centre = radius
2 2
b a
a b c
2
-
= + -
Þ (b – a)
2
= 2(a
2
+ b
2
– c)
Þ (a + b)
2
= 2c
Þ
2
(a b)
1
2c
+
=
4. 1/3
GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
Target : JEE (Main + Advanced) 2021
GR_PT (FOR JEE-ADV.) # 09 (Straight line + Circle) MATHEMATICS
SOLUTION
SECTION–I(i)
1. Ans. (B)
2
3
2 2
2
a + a +
=
2
3 4
a + a + =
2
1 0
a + a - = Þ
1 5
2
+
a = - or
1 5
2
-
-
a has only one value in first quadrant.
2. Ans. (B)
Family of lines can be represented as
a(x + 2y – 5) + b(3x – y – 1) = 0
Hence the family always pass through the point
(1,2).
Equation of other bisector must be of the form
4x + 3y = k. (Q angle bisectors are
perpendicular to each other)
As it pass through the point (1, 2) we get k =
10.
3. Ans. (D)
Area will be maximum when the line x + y + 1
= 0 is the diameter of the given circle.
Hence g
2
+ ƒ
2
= 1
Hence (g, ƒ ) can take the values (0, ±1) or (±1,
0)
4. Ans. (B)
We have, a + 2h + b = 0
Compairing with ax + by = h we get (x, y)
as
1 1
,
2 2
æ ö
- -
ç ÷
è ø
5. Ans. (A)
L1
: (x – y – 6) + l (2x + y + 3)
Always passes through A(1, –5)
and
L2
: (x + 2y – 4) + m(–3x + 2y + 4) = 0
Always passes through B(2, 1)
Since lines intersect at right angle so locus of
point of intersection of L1
and L2
will be a circle
with diametric end points A and B.
Þ x2
+ y2
– 3x + 4y – 3 = 0
6. Ans. (A,B,C,D)
For equilateral D
2 2 2
1 2 3 1 2 2 3 3 1
z z z z z z z z z
+ + = + +
(A) Which is satisfied by –z1– z2 & z3 also
similarly |z1| = |z2| = |z3| = 1
(B) z1 + 1, z2 + 1, z3 + 1 are also satisfying
the
relation
(C)
z1
z2
z3
z +z
2
1 3
z +z
2
1 2
z +z
2
2 3
are also vertices of equilateral D.
(D) 3
1 2 z
z z
, ,
2 2 2
are also vertices of an
equilateral triangle.
7. Ans. (A,D)
For ax + by + c = 0 & c = –2a – 3b
Þ a(x – 2) + b(y – 3) = 0
family always passes through (2, 3)
Let equation of tangent is y =
1
mx
m
+
3 = 2m +
1
m
Þ 2m2
– 3m + 1 = 0
(2m – 1)(m – 1) = 0
1
m
2
= & m = 1
Tangents are y =
x
2
2
+ Þ 2y = x + 4
y = x + 1
8. Ans.(A,B,C,D)
ƒ '(x) = 4x3 + kx2 + 4x + k
= (4x + k) (x2 + 1)
(A) If x Î (0,2) then k < ƒ '(x) < 5 (8 + k)
ƒ '(x) > 0 Þ k > 0
(B) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8
(C) If x Î (–2,2)
5(–8 + k) < ƒ '(x) < 5(8 + k)
ƒ '(x) > 0 Þ 5(–8 + k) > 0
or k > 8
5. GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
2/3
Target : JEE (Main + Advanced) 2021
(D) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8
SECTION–I(iii)
Paragraph for Question 9 to 10
Radius of Ci = 1
Radius of C = 2 1
+
C2 C1
C3
C4
C
9. Ans. (B)
Required probability
( )
2
4.
. 2 1
p
=
p +
( )
4
4 3 2 2
3 2 2
= = -
+
10. Ans. (C)
Let the hyperbola be
2 2
2 2
x y
1
a 2a
- =
Passing through ( ) ( )
( )
2 1 , 2 1
± + ± +
Þ ( )
2
2
2 1 2a
+ =
lengths of L.R =
2
2.2a
a
2 1
4
2
æ ö
+
= ç ÷
ç ÷
è ø
Paragraph for Question 11 to 13
11. Ans.(D)
(–2,0)
A
1
0 (2,0)
S1
B
X
S2
Y
1
(x2
+ y2
+ 3)2
– (4x)2
= 0
Þ S1
: x2
+ y2
– 4x + 3 = 0
S2
: x2
+ y2
+ 4x + 3 = 0
Maximumvalueof
(x1
– x2
)2
+ (y1
– y2
)2
is correspondingto points A
& B
AB = 6
Þ (AB)2
= 36
12. Ans.(B)
q=30°
q=30°
1
(2,0)
(0,0)
1
sin 30
2
q = Þ q = °
Þ Angle = 60°
13. Ans.. (D)
Paragraph for Question 14 to 16
Foot of perpendicular from focus on any of the
tangent always lies on tangent at vertex
h 1 k
,
2 2
+
æ ö
ç ÷
è ø
lies on x = 0
Þ h = – 1
Þ locus is x = –1
P
(h,k) L
S(1,0)
A
(at ,2at)
2
14. Ans. (D)
x = –1
15. Ans. (B)
k
0,
2
æ ö
ç ÷
è ø
also lies on L & tangent at vertex
yt = x + t
2
A º (0, t) Þ k = 2t
(h, k) º (–1, 2t)
y = 3x 4 3 3
+ + is tangent to circle
Þ Perpendicular distance = radius
2
2t 3 4 3 3
1 t
2
+ - -
= +
Þ t 3
= Þ radius = 2
16. Ans. (C)
2
1 t 4
+ = Þ t 15
=
Slope of the tangent Þ
1 1
t 15
=
SECTION–III(i)
1. Ans. 1
1
n 1 2
n
I
0 II
I x .x 1 x dx
-
= -
ò 1
4
24
3
Applying By parts
1 1
n 1 2 3/2 n 2 2 3/2
n
0
0
x (1 x ) (n 1)x (1 x )
I dx
3 3
- -
æ ö
- - -
= - +
ç ÷
è ø
ò
1
n 2 2 2
n
0
(n 1)
I x (1 x ) 1 x dx
3
-
-
= - -
ò
6. 3/3
GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
Target : JEE (Main + Advanced) 2021
n n 2 n
3I (n 1)I (n 1)I
-
= - - - Þ n
n 2
I n 1
I n 2
-
-
=
+
2. Ans. 3
Let eqution of circle be
(x2 + y2 + 2x – 4y – 4) + l(2x – 3y)
Its centre
3
1 ,2
2
l
æ ö
- -l +
ç ÷
è ø
Centre is at minimum distance from (–5,8)
Þ Centre = (–5,8) Þ l = 4
Hence equation of circle is
x2 + y2 + 10x – 16y – 4 = 0
radius 93
=
N
3
31
=
3. Ans. 4
2 5
Ö 90º
B
Ö5
D
A(3,1)
C
ÐABC = 90º
BC 2 5
=
AB 9 1 5 5
= + - =
Þ AC = 5
Power of point Þ AB2
= AD.AC
Þ AD = 1 Þ CD = 4
SECTION–IV
1. Ans. (A)®(R); (B)®(P,Q,R,S); (C)®
(P,Q,S); (D)®(T)
(A) (5!6!.....9!) (10!....14!) (10!...19!)
(20!....24!) (25!.....29!)
In 5!,6!.........,9! exponent of 5 is one.
10!.........14! exponent of 5 is two.
15!.........19! exponent of 5 is 3.
20!.........24! exponent of 5 is 4.
25!.........29! exponent of 5 is 6.
hence it is
5
5
5
10
5
15
5
20
5
30
= 5
80
(B)
( )
2
x
y
1 x
=
+
( )
n n
2
2 2
n 0
|x|dx xdx
2 n 1 n
1 x (1 x )
-
= = +
+ +
ò ò l
A(n) = ln(1 + n
2
) &
2
2
1 1 n
A n
n n
æ ö
+
æ ö
= ç ÷
ç ÷
è ø è ø
l
Þ A(n) –
1
A 2 nn
n
æ ö
= +
ç ÷
è ø
l
Þ
1
A(n) A
n
n 2n
nn
æ ö
- ç ÷
è ø =
l
(even integer)
(C) 2
2
g(x)
(ƒ(x))
= Þ 3
4
g'(x) ƒ'(x)
(ƒ(x))
= -
Þ
2
3 4
ƒ''(x) 3(ƒ'(x))
g''(x) 4
(ƒ(x)) (ƒ(x))
æ ö
= - -
ç ÷
ç ÷
è ø
Þ
3 2
3 4
ƒ''(x) 3(g'(x)(ƒ(x) )
g''(x) 4
(ƒ(x)) 16ƒ(x)
æ ö
= - -
ç ÷
ç ÷
è ø
=
2 2
3
ƒ''(x) 3
4 (g'(x)) (ƒ(x))
16
(ƒ(x))
æ ö
- -
ç ÷
è ø
l = 3
(D) Radical axis of both the circles is x – y = 0
Length of perpendicular from
centre = radius
2 2
b a
a b c
2
-
= + -
Þ (b – a)
2
= 2(a
2
+ b
2
– c)
Þ (a + b)
2
= 2c
Þ
2
(a b)
1
2c
+
=
7. 1/3
GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
Target : JEE (Main + Advanced) 2021
GR_PT (FOR JEE-ADV.) # 09 (Straight line + Circle) MATHEMATICS
SOLUTION
SECTION–I(i)
1. Ans. (B)
2
3
2 2
2
a + a +
=
2
3 4
a + a + =
2
1 0
a + a - = Þ
1 5
2
+
a = - or
1 5
2
-
-
a has only one value in first quadrant.
2. Ans. (B)
Family of lines can be represented as
a(x + 2y – 5) + b(3x – y – 1) = 0
Hence the family always pass through the point
(1,2).
Equation of other bisector must be of the form
4x + 3y = k. (Q angle bisectors are
perpendicular to each other)
As it pass through the point (1, 2) we get k =
10.
3. Ans. (D)
Area will be maximum when the line x + y + 1
= 0 is the diameter of the given circle.
Hence g
2
+ ƒ
2
= 1
Hence (g, ƒ ) can take the values (0, ±1) or (±1,
0)
4. Ans. (B)
We have, a + 2h + b = 0
Compairing with ax + by = h we get (x, y)
as
1 1
,
2 2
æ ö
- -
ç ÷
è ø
5. Ans. (A)
L1
: (x – y – 6) + l (2x + y + 3)
Always passes through A(1, –5)
and
L2
: (x + 2y – 4) + m(–3x + 2y + 4) = 0
Always passes through B(2, 1)
Since lines intersect at right angle so locus of
point of intersection of L1
and L2
will be a circle
with diametric end points A and B.
Þ x2
+ y2
– 3x + 4y – 3 = 0
6. Ans. (A,B,C,D)
For equilateral D
2 2 2
1 2 3 1 2 2 3 3 1
z z z z z z z z z
+ + = + +
(A) Which is satisfied by –z1– z2 & z3 also
similarly |z1| = |z2| = |z3| = 1
(B) z1 + 1, z2 + 1, z3 + 1 are also satisfying
the
relation
(C)
z1
z2
z3
z +z
2
1 3
z +z
2
1 2
z +z
2
2 3
are also vertices of equilateral D.
(D) 3
1 2 z
z z
, ,
2 2 2
are also vertices of an
equilateral triangle.
7. Ans. (A,D)
For ax + by + c = 0 & c = –2a – 3b
Þ a(x – 2) + b(y – 3) = 0
family always passes through (2, 3)
Let equation of tangent is y =
1
mx
m
+
3 = 2m +
1
m
Þ 2m2
– 3m + 1 = 0
(2m – 1)(m – 1) = 0
1
m
2
= & m = 1
Tangents are y =
x
2
2
+ Þ 2y = x + 4
y = x + 1
8. Ans.(A,B,C,D)
ƒ '(x) = 4x3 + kx2 + 4x + k
= (4x + k) (x2 + 1)
(A) If x Î (0,2) then k < ƒ '(x) < 5 (8 + k)
ƒ '(x) > 0 Þ k > 0
(B) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8
(C) If x Î (–2,2)
5(–8 + k) < ƒ '(x) < 5(8 + k)
ƒ '(x) > 0 Þ 5(–8 + k) > 0
or k > 8
8. GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
2/3
Target : JEE (Main + Advanced) 2021
(D) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8
SECTION–I(iii)
Paragraph for Question 9 to 10
Radius of Ci = 1
Radius of C = 2 1
+
C2 C1
C3
C4
C
9. Ans. (B)
Required probability
( )
2
4.
. 2 1
p
=
p +
( )
4
4 3 2 2
3 2 2
= = -
+
10. Ans. (C)
Let the hyperbola be
2 2
2 2
x y
1
a 2a
- =
Passing through ( ) ( )
( )
2 1 , 2 1
± + ± +
Þ ( )
2
2
2 1 2a
+ =
lengths of L.R =
2
2.2a
a
2 1
4
2
æ ö
+
= ç ÷
ç ÷
è ø
Paragraph for Question 11 to 13
11. Ans.(D)
(–2,0)
A
1
0 (2,0)
S1
B
X
S2
Y
1
(x2
+ y2
+ 3)2
– (4x)2
= 0
Þ S1
: x2
+ y2
– 4x + 3 = 0
S2
: x2
+ y2
+ 4x + 3 = 0
Maximumvalueof
(x1
– x2
)2
+ (y1
– y2
)2
is correspondingto points A
& B
AB = 6
Þ (AB)2
= 36
12. Ans.(B)
q=30°
q=30°
1
(2,0)
(0,0)
1
sin 30
2
q = Þ q = °
Þ Angle = 60°
13. Ans.. (D)
Paragraph for Question 14 to 16
Foot of perpendicular from focus on any of the
tangent always lies on tangent at vertex
h 1 k
,
2 2
+
æ ö
ç ÷
è ø
lies on x = 0
Þ h = – 1
Þ locus is x = –1
P
(h,k) L
S(1,0)
A
(at ,2at)
2
14. Ans. (D)
x = –1
15. Ans. (B)
k
0,
2
æ ö
ç ÷
è ø
also lies on L & tangent at vertex
yt = x + t
2
A º (0, t) Þ k = 2t
(h, k) º (–1, 2t)
y = 3x 4 3 3
+ + is tangent to circle
Þ Perpendicular distance = radius
2
2t 3 4 3 3
1 t
2
+ - -
= +
Þ t 3
= Þ radius = 2
16. Ans. (C)
2
1 t 4
+ = Þ t 15
=
Slope of the tangent Þ
1 1
t 15
=
SECTION–III(i)
1. Ans. 1
1
n 1 2
n
I
0 II
I x .x 1 x dx
-
= -
ò 1
4
24
3
Applying By parts
1 1
n 1 2 3/2 n 2 2 3/2
n
0
0
x (1 x ) (n 1)x (1 x )
I dx
3 3
- -
æ ö
- - -
= - +
ç ÷
è ø
ò
1
n 2 2 2
n
0
(n 1)
I x (1 x ) 1 x dx
3
-
-
= - -
ò
9. 3/3
GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
Target : JEE (Main + Advanced) 2021
n n 2 n
3I (n 1)I (n 1)I
-
= - - - Þ n
n 2
I n 1
I n 2
-
-
=
+
2. Ans. 3
Let eqution of circle be
(x2 + y2 + 2x – 4y – 4) + l(2x – 3y)
Its centre
3
1 ,2
2
l
æ ö
- -l +
ç ÷
è ø
Centre is at minimum distance from (–5,8)
Þ Centre = (–5,8) Þ l = 4
Hence equation of circle is
x2 + y2 + 10x – 16y – 4 = 0
radius 93
=
N
3
31
=
3. Ans. 4
2 5
Ö 90º
B
Ö5
D
A(3,1)
C
ÐABC = 90º
BC 2 5
=
AB 9 1 5 5
= + - =
Þ AC = 5
Power of point Þ AB2
= AD.AC
Þ AD = 1 Þ CD = 4
SECTION–IV
1. Ans. (A)®(R); (B)®(P,Q,R,S); (C)®
(P,Q,S); (D)®(T)
(A) (5!6!.....9!) (10!....14!) (10!...19!)
(20!....24!) (25!.....29!)
In 5!,6!.........,9! exponent of 5 is one.
10!.........14! exponent of 5 is two.
15!.........19! exponent of 5 is 3.
20!.........24! exponent of 5 is 4.
25!.........29! exponent of 5 is 6.
hence it is
5
5
5
10
5
15
5
20
5
30
= 5
80
(B)
( )
2
x
y
1 x
=
+
( )
n n
2
2 2
n 0
|x|dx xdx
2 n 1 n
1 x (1 x )
-
= = +
+ +
ò ò l
A(n) = ln(1 + n
2
) &
2
2
1 1 n
A n
n n
æ ö
+
æ ö
= ç ÷
ç ÷
è ø è ø
l
Þ A(n) –
1
A 2 nn
n
æ ö
= +
ç ÷
è ø
l
Þ
1
A(n) A
n
n 2n
nn
æ ö
- ç ÷
è ø =
l
(even integer)
(C) 2
2
g(x)
(ƒ(x))
= Þ 3
4
g'(x) ƒ'(x)
(ƒ(x))
= -
Þ
2
3 4
ƒ''(x) 3(ƒ'(x))
g''(x) 4
(ƒ(x)) (ƒ(x))
æ ö
= - -
ç ÷
ç ÷
è ø
Þ
3 2
3 4
ƒ''(x) 3(g'(x)(ƒ(x) )
g''(x) 4
(ƒ(x)) 16ƒ(x)
æ ö
= - -
ç ÷
ç ÷
è ø
=
2 2
3
ƒ''(x) 3
4 (g'(x)) (ƒ(x))
16
(ƒ(x))
æ ö
- -
ç ÷
è ø
l = 3
(D) Radical axis of both the circles is x – y = 0
Length of perpendicular from
centre = radius
2 2
b a
a b c
2
-
= + -
Þ (b – a)
2
= 2(a
2
+ b
2
– c)
Þ (a + b)
2
= 2c
Þ
2
(a b)
1
2c
+
=
10. 1/3
GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
Target : JEE (Main + Advanced) 2021
GR_PT (FOR JEE-ADV.) # 09 (Straight line + Circle) MATHEMATICS
SOLUTION
SECTION–I(i)
1. Ans. (B)
2
3
2 2
2
a + a +
=
2
3 4
a + a + =
2
1 0
a + a - = Þ
1 5
2
+
a = - or
1 5
2
-
-
a has only one value in first quadrant.
2. Ans. (B)
Family of lines can be represented as
a(x + 2y – 5) + b(3x – y – 1) = 0
Hence the family always pass through the point
(1,2).
Equation of other bisector must be of the form
4x + 3y = k. (Q angle bisectors are
perpendicular to each other)
As it pass through the point (1, 2) we get k =
10.
3. Ans. (D)
Area will be maximum when the line x + y + 1
= 0 is the diameter of the given circle.
Hence g
2
+ ƒ
2
= 1
Hence (g, ƒ ) can take the values (0, ±1) or (±1,
0)
4. Ans. (B)
We have, a + 2h + b = 0
Compairing with ax + by = h we get (x, y)
as
1 1
,
2 2
æ ö
- -
ç ÷
è ø
5. Ans. (A)
L1
: (x – y – 6) + l (2x + y + 3)
Always passes through A(1, –5)
and
L2
: (x + 2y – 4) + m(–3x + 2y + 4) = 0
Always passes through B(2, 1)
Since lines intersect at right angle so locus of
point of intersection of L1
and L2
will be a circle
with diametric end points A and B.
Þ x2
+ y2
– 3x + 4y – 3 = 0
6. Ans. (A,B,C,D)
For equilateral D
2 2 2
1 2 3 1 2 2 3 3 1
z z z z z z z z z
+ + = + +
(A) Which is satisfied by –z1– z2 & z3 also
similarly |z1| = |z2| = |z3| = 1
(B) z1 + 1, z2 + 1, z3 + 1 are also satisfying
the
relation
(C)
z1
z2
z3
z +z
2
1 3
z +z
2
1 2
z +z
2
2 3
are also vertices of equilateral D.
(D) 3
1 2 z
z z
, ,
2 2 2
are also vertices of an
equilateral triangle.
7. Ans. (A,D)
For ax + by + c = 0 & c = –2a – 3b
Þ a(x – 2) + b(y – 3) = 0
family always passes through (2, 3)
Let equation of tangent is y =
1
mx
m
+
3 = 2m +
1
m
Þ 2m2
– 3m + 1 = 0
(2m – 1)(m – 1) = 0
1
m
2
= & m = 1
Tangents are y =
x
2
2
+ Þ 2y = x + 4
y = x + 1
8. Ans.(A,B,C,D)
ƒ '(x) = 4x3 + kx2 + 4x + k
= (4x + k) (x2 + 1)
(A) If x Î (0,2) then k < ƒ '(x) < 5 (8 + k)
ƒ '(x) > 0 Þ k > 0
(B) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8
(C) If x Î (–2,2)
5(–8 + k) < ƒ '(x) < 5(8 + k)
ƒ '(x) > 0 Þ 5(–8 + k) > 0
or k > 8
11. GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
2/3
Target : JEE (Main + Advanced) 2021
(D) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8
SECTION–I(iii)
Paragraph for Question 9 to 10
Radius of Ci = 1
Radius of C = 2 1
+
C2 C1
C3
C4
C
9. Ans. (B)
Required probability
( )
2
4.
. 2 1
p
=
p +
( )
4
4 3 2 2
3 2 2
= = -
+
10. Ans. (C)
Let the hyperbola be
2 2
2 2
x y
1
a 2a
- =
Passing through ( ) ( )
( )
2 1 , 2 1
± + ± +
Þ ( )
2
2
2 1 2a
+ =
lengths of L.R =
2
2.2a
a
2 1
4
2
æ ö
+
= ç ÷
ç ÷
è ø
Paragraph for Question 11 to 13
11. Ans.(D)
(–2,0)
A
1
0 (2,0)
S1
B
X
S2
Y
1
(x2
+ y2
+ 3)2
– (4x)2
= 0
Þ S1
: x2
+ y2
– 4x + 3 = 0
S2
: x2
+ y2
+ 4x + 3 = 0
Maximumvalueof
(x1
– x2
)2
+ (y1
– y2
)2
is correspondingto points A
& B
AB = 6
Þ (AB)2
= 36
12. Ans.(B)
q=30°
q=30°
1
(2,0)
(0,0)
1
sin 30
2
q = Þ q = °
Þ Angle = 60°
13. Ans.. (D)
Paragraph for Question 14 to 16
Foot of perpendicular from focus on any of the
tangent always lies on tangent at vertex
h 1 k
,
2 2
+
æ ö
ç ÷
è ø
lies on x = 0
Þ h = – 1
Þ locus is x = –1
P
(h,k) L
S(1,0)
A
(at ,2at)
2
14. Ans. (D)
x = –1
15. Ans. (B)
k
0,
2
æ ö
ç ÷
è ø
also lies on L & tangent at vertex
yt = x + t
2
A º (0, t) Þ k = 2t
(h, k) º (–1, 2t)
y = 3x 4 3 3
+ + is tangent to circle
Þ Perpendicular distance = radius
2
2t 3 4 3 3
1 t
2
+ - -
= +
Þ t 3
= Þ radius = 2
16. Ans. (C)
2
1 t 4
+ = Þ t 15
=
Slope of the tangent Þ
1 1
t 15
=
SECTION–III(i)
1. Ans. 1
1
n 1 2
n
I
0 II
I x .x 1 x dx
-
= -
ò 1
4
24
3
Applying By parts
1 1
n 1 2 3/2 n 2 2 3/2
n
0
0
x (1 x ) (n 1)x (1 x )
I dx
3 3
- -
æ ö
- - -
= - +
ç ÷
è ø
ò
1
n 2 2 2
n
0
(n 1)
I x (1 x ) 1 x dx
3
-
-
= - -
ò
12. 3/3
GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION
Target : JEE (Main + Advanced) 2021
n n 2 n
3I (n 1)I (n 1)I
-
= - - - Þ n
n 2
I n 1
I n 2
-
-
=
+
2. Ans. 3
Let eqution of circle be
(x2 + y2 + 2x – 4y – 4) + l(2x – 3y)
Its centre
3
1 ,2
2
l
æ ö
- -l +
ç ÷
è ø
Centre is at minimum distance from (–5,8)
Þ Centre = (–5,8) Þ l = 4
Hence equation of circle is
x2 + y2 + 10x – 16y – 4 = 0
radius 93
=
N
3
31
=
3. Ans. 4
2 5
Ö 90º
B
Ö5
D
A(3,1)
C
ÐABC = 90º
BC 2 5
=
AB 9 1 5 5
= + - =
Þ AC = 5
Power of point Þ AB2
= AD.AC
Þ AD = 1 Þ CD = 4
SECTION–IV
1. Ans. (A)®(R); (B)®(P,Q,R,S); (C)®
(P,Q,S); (D)®(T)
(A) (5!6!.....9!) (10!....14!) (10!...19!)
(20!....24!) (25!.....29!)
In 5!,6!.........,9! exponent of 5 is one.
10!.........14! exponent of 5 is two.
15!.........19! exponent of 5 is 3.
20!.........24! exponent of 5 is 4.
25!.........29! exponent of 5 is 6.
hence it is
5
5
5
10
5
15
5
20
5
30
= 5
80
(B)
( )
2
x
y
1 x
=
+
( )
n n
2
2 2
n 0
|x|dx xdx
2 n 1 n
1 x (1 x )
-
= = +
+ +
ò ò l
A(n) = ln(1 + n
2
) &
2
2
1 1 n
A n
n n
æ ö
+
æ ö
= ç ÷
ç ÷
è ø è ø
l
Þ A(n) –
1
A 2 nn
n
æ ö
= +
ç ÷
è ø
l
Þ
1
A(n) A
n
n 2n
nn
æ ö
- ç ÷
è ø =
l
(even integer)
(C) 2
2
g(x)
(ƒ(x))
= Þ 3
4
g'(x) ƒ'(x)
(ƒ(x))
= -
Þ
2
3 4
ƒ''(x) 3(ƒ'(x))
g''(x) 4
(ƒ(x)) (ƒ(x))
æ ö
= - -
ç ÷
ç ÷
è ø
Þ
3 2
3 4
ƒ''(x) 3(g'(x)(ƒ(x) )
g''(x) 4
(ƒ(x)) 16ƒ(x)
æ ö
= - -
ç ÷
ç ÷
è ø
=
2 2
3
ƒ''(x) 3
4 (g'(x)) (ƒ(x))
16
(ƒ(x))
æ ö
- -
ç ÷
è ø
l = 3
(D) Radical axis of both the circles is x – y = 0
Length of perpendicular from
centre = radius
2 2
b a
a b c
2
-
= + -
Þ (b – a)
2
= 2(a
2
+ b
2
– c)
Þ (a + b)
2
= 2c
Þ
2
(a b)
1
2c
+
=
