a2
a2
a) ⎯
⎯
√5
b) ⎯
⎯
√7
a2
d) ⎯
⎯
√2
Sol: Given: x2+y2−2ax = 0
C (a, 0)
d) None
Sol: Given circle = x2 + y2 − 8x − 12y + p = 0
⎯
⎯
⎯
⎯
√ 16 + 36 − p
C = (4, 6); r =
√ 52 − p
=
1) r < 4
Equation of line
through C parallel to
x+2y = 0 is
i.e:
x+2y−a = 0
36 < p
AB = diameter of the
circle = 2a
2) r < 6
a
Length of perpendicular from O upon the line = ⎯
⎯
√5
1
a
a2
∴ Area of Δ AOB = ⎯ × 2a × ⎯
⎯ = ⎯
⎯ .
2
√5
√5
√
(4, 6)
⎯
⎯
⎯
52 − p
< 4
52 − p < 16
52 −16 < p
√
x2+y2−2x−7 = 0,
C
=
⎯
(1,
0);
⎯
√1+7 = 2√2
x2 + y2 - 2x - 15 = 0
([p+1], [p] )
= ([p] + 1, [p])
( [p] + 1, [p] ( lies
2x - 7 =
y2 2 +
x2 + y2 − 2x − 15 = 0
⎯
⎯
⎯
i.e: 52 − p
C = (1, 0), r = 4
r=
Conditions:
C = (a, 0); radius = a.
ISEET - JEE
Maths - Circle
Sol: x2+y2−2x−15 = 0;
< 6
52 − p < 36
16 < p
3) S11 < 0
4 + 25 −16 −60 + p < 0
2[p]2 − 16 < 0
29 − 76 + p < 0
[p]2 − 8 < 0
p < 47
[p]2 < 8 ....... (1)
([p] + 1, [p] ) lies
x2 + y2 − 2x − 7 = 0
[p]2 + 1 + 2[p] + [p]2 −2[p] − 9 > 0
2[p]2 − 8 > 0
Mathematics
b) (x−15)2 + (y−14)2 = 1
[p]2 > 4 ....... (2)
c) (x−16)2 + (y−15)2 = 1
from (1) & (2)
d) (x−17)2 + (y−16)2 = 1
4 < [p]2 < 8.
x = a are
Sol:
b) x2 + y2 + 4x − 2 = 0
(
)
a -−
a
−
,
2
2
7. A triangle is formed by the lines whose
combined equation is given by
(x + y − 4) (xy − 2x − y + 2) = 0. The
c) x2 + y2 + 4x + 2 = 0
d) None
=
y
{
c) x2 + y2 + 2x + 2y − 3 = 0
r
m
(0, 2)
y
=
4
(1, 3)
√1 + 4 + 2 √3⎯−1
radius = R =
√
=
⎯⎯
⎯
⎯
⎯
4 + 2√3 = (1+√3)
Let r be the radius of required circle
R = PC + PA
R − r = PC =
(2, 0)
(1, 0)
A
⎯
⎯
⎯⎯
√ (x1−1)2 + (y1−2)2
P (x1, y1)
r
r
60°
→ x+y = 1 (... n = 1)
A
d = Perpendicular distance from
1
(0, 0) to the chord = ⎯
⎯
√2
⎯
⎯
→ x + y = 2 ( ... n = 2)
=
⎯ 3
⇒ (1+ √3 ) = ⎯
2
⎯
2
⎯ (1+ √3 ) =
3
⎯
⎯
⎯⎯
2 + (y −2)2
1
√ (x1−1)
⎯⎯
√ (x1−1)2 + (y1−2)2
S.B.S.
4
⎯
⎯ (1+ √3 )2 = (x1−1)2 + (y1−2)2
9
⎯
√2
(0, 2)
Locus is (x−1)2 + (y−2)2 =
=
−x
(0, 2)
⇒ (x − 2)2 + y2 = 2
⇒ x2 + y2 − 4x + 2 = 0.
4. If the circle x2 + y2 + 4x + 22y + c = 0 bisects
the
circumference
of
the
circle
x2 + y2 − 2x + 8y − d = 0, then c + d is equal to
a) 60
b) 50
c) 40
d) 56
Sol: x2 + y2 + 4x + 22y + c = 0 bisects the circumference of the circle
x2 + y2 − 2x + 8y − d = 0 means that radical
axes is the diameter of
x2 + y2 − 2x + 8y − d = 0
Radical axes: 6x + 14y + c + d = 0
Centre = (1, −4) lies on radical axes
6 − 56 + c + d = 0
∴ c + d = 50
5. If (2, 5) is an interior point of the circle
x2 + y2 − 8x − 12y + p = 0 and the circle
(0, 1)
(2, 2)
(2, 0)
(1, 0)
(1, 2)
x+y=2
x+y=1
(1, 0)
⎯
⎯
l2 = length of chord = 2 √ 4 - 2 = 2√ 2
(x + y − 4) (x(y − 2)−1(y − 2)) = 0
(x + y − 4) (x − 1) (y − 2) = 0
x + y − 4 = 0; x = 1; y = 2
(x − 1) (x − 2) + (y − 3) (y − 2) = 0
⎯
a) 4− √2
⎯
c) 4 + 2 √2
b) 6
d) 4 +
Since the equation of OP is x+y = 0
⎯
∴OP = 2√2 = CP
10. The area bounded by the circles x2+y2 = 1,
Hence OC = 4
√ 3(x2 + y2) = 4xy, is equal to ......
the circle with
⎯
5Π
b) ⎯
2
c) 3Π
Π
d) ⎯
4
4
Sol: x2+y2 = 1, x2+y2 = 4, x2 + y2 − ⎯
⎯ xy = 0
√3
a = 1, b = 1,
.
⎯
then ∠COP = 45°
Π
a) ⎯
2
)
√2
7
= l12 + l22 = 4. ⎯ + 8 = 22
2
⇒ x2 +y2 − 3x − 5y + 8 = 0
a) p ∈ [−1 0) ∪ (0 1) ∪ (1, 2)
3
Sol: If (a, 0) is the centre C and P is (2, −2).
The point on
inside the region bounded by the circle x2 +
y2 − 2x − 15 = 0 and x2 + y2 − 2x − 7 = 0 then
(
2
⎯
Sum of squares
x2+y2 = 4 and the pair of lines
( [p + 1], [p] ) (Where [x] is the
greatest integer less than or equal to x), lies
⎯
2(1+ √3 )
12. If (α, β) is a point on the circle whose centre is on the X−axis and which touches the
line x + y = 0 at (2, −2), then the greatest
value of α is ....
⇒ x2 + y2 − 3x − 5y + 2 + 6 = 0
8. If the point
°
30° 30
2)
C (1,
y
(1, 2)
r
, y 1)
P (x 1
r
r
60°
C (1, 2)
y=2
∴ Required equation is
2
(x−2)2+ (y − 0)2 = (√⎯
2)
⎯
√3−1) = 0
PA
Now, sin30° = ⎯
PC
(0, 1)
⏐0 + 0 − 2⏐
d = ⎯⎯
⎯
√ 1+1
x=1
+
{
(1/√⎯2 ) (1/√⎯2 )
d) None
Sol:
C(2, 0)
d) None
1 = 2 √−
7
l1 = length of chord = 2 √ 4 − −
2
2
b) x2 + y2 −3x − 5y + 8 = 0
x
⇒ r = √⎯
⎯
2
2
r (0, 0) √
sin 45° = ⎯
OC
O
OC =
⎯
√2 ⎯
r
√2
⎯=⎯ =2
equation of its circumcircle is ......
a) x2 + y2 − 5x − 3y + 8 = 0
x
r
Sol: tan 45° = ⎯
⎯
√2
c) 33
Sol: Centre = (0, 0), r = 2
di
a2
x2+y2+ax ± ay+ ⎯ = 0
4
.co
a
−
2
⎯
b) (x − 2)2 + (y − 1)2 = 1 + 2 √3
c) x2 + y2 = 1
d) None of these
⎯
⎯
⎯⎯
The sum of the squares of the
lengths of the chords intercepted
b) 22
a) (x − 1)2 + (y − 2)2 = 3
Sol: Given x2 + y2 − 2x − 4y −(2
by the line x+y = n, n ∈ N on the circle
x2 + y2 = 4 is
a) 11
in
a) x2 + y2 − 4x + 2 = 0
a
−
2
a2
∴ Required circle is (x−17)2 + (y−16)2 = 1
3. The equation of the circle touching the
lines⎥ y⎥ = x at a distance √⎯
2 units from the
origin is ...........
(−a2, −a2)
a2
b) x2+y2+ax ± ay+ ⎯ = 0
4
α = 17, β = 16
α − 3 = β − 2 = 14
x=a
a2
a) x2+y2 ± ax ± ay+ ⎯ = 0
4
c) x2+y2−ax ± ay+ ⎯ = 0
4
d) None
(α, β)
9.
id
ya
−2(3 + 2 − 19)
= ⎯⎯
12 + 12
6. The equations of the circle which
touch both the axes and the line
Π
Π
= − (3) = −
12
4
centre = (1, 2)
av
Sol: The image of the circle has same radius but
centre is different.
(3, 2)
If centre is (α , β), then
19
α−3
β−2
y=
⎯ =⎯
+
x
1
1
Π/6
11. The locus of the centre of a circle touching
⎯
the circle x2+y2−4y−2x = 2 √3 − 1 internally
and tangents on which from (1, 2) is making
a 60° angle with each other, is
([p] + 1)2 + [p]2 − 2[p] − 2 − 7 > 0
a) (x−14)2 + (y−13)2 = 1
⎯
√ (a-b)2 + 4h2
( )
ISEET - IIT JEE Study Material
Mathematics
2. The equation of the image of the circle
(x−3)2 +(y−2)2 =1 by the mirror x + y = 19 is
a +b
cosθ = ⎯⎯
1
2
2
= ⎯⎯
⎯
√ 02 + 4 −34
⎯
2 √3
= ⎯⎯ ⇒ θ = 30°
4
1
Π
1
Π
Required area = −(2)2. − − − (1)2 −
2
6
2
6
C(1, 0)
[p]2+1+2[p] + [p]2
− 2[p] − 2 − 15 < 0
∴ p ∈ (16, 47)
4
2h = − ⎯
⎯
√3
>
B
c) p ∈ (16, 36)
c) p ∈ (−1 2)
d) None
0
O
(0, 0)
b) p ∈ (16, 47)
a2
c) ⎯
⎯
√3
A
a) p ∈ (36, 47)
b) p ∈ (0 1)
<
1. A straight line is drawn through the centre
x2+y2−2ax = 0 parallel to the straight line
x+2y = 0 and intersecting the circle at A and
B. Then the area of Δ AOB is .......
neither cut nor touches any one of the axes
of co-ordinates, then
x
CIRCLE
C(a,0)
A
(0, 0)
P(2, -2)
the greatest
X - co-ordinate is B
B
45°
x+y
=0
⎯
α = OB = OC + CB = 4 + 2√2
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