Test: 35 Year JEE Previous Year Questions: Circle


29 Questions MCQ Test Mathematics For JEE | Test: 35 Year JEE Previous Year Questions: Circle


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Attempt Test: 35 Year JEE Previous Year Questions: Circle | 29 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics For JEE for JEE Exam | Download free PDF with solutions
QUESTION: 1

If the chord y = mx + 1 of the circle x2+y2=1 subtends an angle of measure 45° at the major segment of the circle then value of m is [2002]

Solution:

Equation of circle x2 + y2 = 1 = (1)2
⇒ x2 + y2 = (y – mx)2 ⇒ x2 = m2x2 – 2 mxy;
⇒ x2 (1 – m2) + 2mxy = 0.
Which represents the pair of lines between which the angle is 45o.

⇒ 1 – m2 = ± 2m ⇒ m2 ± 2m – 1 = 0

QUESTION: 2

The centres of a set of circles, each of radius 3, lie on the circle x2+y2=25. The locus of any point in the set is

Solution:

For any point P (x, y) in the given circle,

we should have

OA ≤ OP ≤ OB ⇒ (5-3)

⇒ 4 ≤ x2 +y2 ≤ 64

QUESTION: 3

The centre of the circle passing through (0, 0) and (1, 0) and touching the circle x2+y2=9 is [2002]

Solution:

Let the required circle be x2  +  y2 + 2gx  + 2fy + c = 0
Since it passes through (0, 0) and (1, 0)

⇒ c = 0 and  g = -

Points (0, 0) and (1, 0) lie inside the circle x2 + y2 = 9, so two circles touch internally
⇒ c1c2 = r1 – r2

Hence, the centres of required circle are

QUESTION: 4

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is

Solution:

Let ABC be an equilateral triangle, whose median is AD.

Given AD = 3a.
In Δ ABD, AB2 = AD2 + BD2 ; ⇒ x2 = 9a2 + (x2/4)  

where AB = BC = AC = x.

In Δ OBD, OB2 = OD2 + BD2

⇒ r2 = 9a2 – 6ar + r2 + 3a2 ;
⇒ 6ar = 12a2 ⇒ r = 2a
So equation of circle is x2 + y2 = 4a2

QUESTION: 5

If the two circles ( x - 1)2 + ( y - 3)2 = r2 and x2 + y2 - 8 x + 2 y + 8 = 0 intersect in two distinct point, then[2003]

Solution:

r1 - r2< C1C2 for intersection

and ,
From (1) and (2),  2 < r < 8.

QUESTION: 6

The lines 2 x - 3y=5 and 3x - 4 y = 7 are diameters of a circle having area as 154 sq.units.Then the equation  of the circle is[2003]

Solution:

πr2 = 154 ⇒r= 7
For centre on solving equation 2x -3y = 5& 3x -4y= 7
we get x = 1,y =-1
∴ centre = (1, –1 )
Equation of circle, ( x - 1)2 + (y + 1)2
=72 x2 + y2 -2x+2y= 47

QUESTION: 7

If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is

Solution:

Let the variable circle is x2 + y2 + 2gx + 2fy + c = 0 ......(1)
It passes through (a, b)
∴ a2 + b2 + 2 ga + 2 fb + c = 0 ......(2)

(1) cuts x2 +y2= 4 orthogonally

∴2(gx0+ f x0) = c-4 ⇒ c=4

∴ from (2) a2 + b2 + 2ga + 2fb + 4=0

∴  Locus of centre (–g,–f) is a2 + b2 - 2ax -2by +4= 0
or 2ax + 2by = a2 + b+ 4

QUESTION: 8

A variable circle passes through the fixed point A( p,q) and touches x-axis . The locus of the other end of the diameter through A is [2004]

Solution:

Let the variable circle be x2 + y2 + 2gx + 2fy + c = 0 ....(1)

∴ p2 + q2 + 2gp + 2fq + c =0 ....(2)
Circle (1) touches x-axis,
∴ g2 - c = 0 ⇒ c=g2 .
From (2) p2 + q2 + 2gp + 2fq +g=0 ....(3)
Let the other end of diameter through (p, q) be (h, k),

then 

Put in (3)

⇒ h2 + p2 - 2hp - 4kq=0

∴ locus of (h, k) is x2 + p2 - 2 xp - 4 yq=0

⇒ ( x - p )2=4qy

QUESTION: 9

If the lines 2x + 3 y + 1 = 0 and 3x -y - 4=0 lie alon g diameter of a circle of circumference 10p, then the equation of the circle is [2004]

Solution:

Two diameters are along 2x + 3y +1 = 0 and 3x -y - 4=0
solving we get centre (1, –1) circumference = 2πr = 10π
∴ r= 5.
Required circle is, ( x - 1)2 + (y + 1)2=52
⇒ x2 + y2 -2x+ 2y -23=0

QUESTION: 10

Intercept on the line y = x by the circle x2 + y2 - 2x = 0 is AB. Equation of the circle on AB as a diameter is [2004]

Solution:

Solving y = x and the circle x2 + y2 - 2x= 0,
we get x = 0, y = 0 and x = 1,y=1
∴ Extremities of diameter of the required circle are (0, 0) and (1, 1).
Hence, the equation of circle is (x - 0)(x -1) + (y - 0)(y -1)= 0
⇒ x2 + y2 - x -y= 0

QUESTION: 11

If th e circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1 = 0 intersect in two distinct points P and Q then the line 5x + by – a = 0 passes through P and Q for[2005]

Solution:

s1 = x2 + y2+ 2ax + cy +a =0  
s2 = x+ y- 3ax + dy - 1 =0
Equation of common chord of circles s1 and s2 is given by s1 - s2= 0
⇒ 5ax + (c - d)y +a +1=0
Given that 5x + by – a = 0
passes through P and Q

∴ The two equations should represent the same line

a2 +a+1=0

No real value of a.

QUESTION: 12

A circle touches the x- axis and also touches the circle with centre at (0,3 ) and radius 2. The locus of the centre of the circle is[2005]

Solution:

Equation of circle with centre (0, 3) and radius  2 is x2 + (y - 3)2 = 4

Let locus of the variable circle is (α , β)

∵ It touches x - axis.

∴ It's equation is ( x - α)2 + (y + β)2 = β2

Circle touch externally ⇒ c1c2 = r+ r2

∴ Locus is which is a parabola.

QUESTION: 13

If a circle passes through the point (a, b) and cuts the circle x2 + y2 = p2 orthogonally, then the equation of the locus of its centre is [2005]

Solution:

Let the centre be (α, β)

∵ It cuts the circle x2 + y= p2 orthogonally
∴ Using 2 g1 g2 +2 f1 f2 = c1+c2 , we get  
2(-α) x 0 + 2(-β) x 0 = c1- p2 ⇒ c1 = p2
Let equation of circle is x2 + y2 - 2αx- 2βy+p2 =0
It passes through (a, b) ⇒ a2 + b2 - 2αa - 2βb +p2=0
∴Locus of (α, β) is
∴2ax+ 2by - (a2 +b2 + p2)=0 .

QUESTION: 14

If the pair of lines ax2 + 2 (a + b)xy + by 2 = 0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then [2005]

Solution:

As per question area of one sector = 3 area of another sector
⇒ angle at centre by one sector = 3x angle at centre by another sector
Let one angle be θ then other = 3θ
Clearly θ + 3θ  = 180 ⇒ θ = 45o

∴ Angle between the diameters represented by combined equation

ax2 + 2 ( a + b ) xy + by= 0 is 45o

∴ Using 

we get 

 

⇒ a2 + b2 + 2ab = 4a2 + 4b+4ab ⇒ 3a+ 3b2 + 2ab = 0

QUESTION: 15

If the lines 3x - 4y - 7=0 and 2x - 3y - 5=0 are two diameters of a circle of area 49π square units, the equation of the circle is [2006]

Solution:

Point of intersection of 3x - 4y - 7=0 and 2 x - 3y - 5=0 is (1, - 1)

which is the centre of the circle and radius = 7

∴ Equation is ( x - 1)2 + (y + 1)2= 49

⇒ x2 + y2 -2x+ 2y-47=0

QUESTION: 16

Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle C that subtend an angle  of  at its center is

Solution:

Let M(h, k) be the mid point of chord AB where

∴ Locus of (h, k) is 

QUESTION: 17

Consider a family of circles which are passing through the point (– 1, 1) and are tangent to x-axis. If (h, k) are the coordinate of the centre of the circles, then the set of values of k is given by the interval [2007]

Solution:

Equation of circle whose centre is (h, k)

i.e (x – h)2 + (y – k)2 = k2

(radius of circle = k because circle is tangent to x-axis) Equation of circle passing through (–1, +1)

∴ (–1 –h)2 + (1 – k)2 = k2

⇒ 1 + h2 + 2h + 1 + k2 – 2k = k2

⇒ h2 + 2h – 2k + 2 = 0 D ≥ 0
∴ (2)2 – 4 × 1.(–2k + 2) ≥ 0 ⇒ 4 – 4(–2k + 2) ≥ 0 ⇒ 1 + 2k – 2 ≥ 0 ⇒

QUESTION: 18

The point diametrically opposite to the point P(1, 0) on the circle x2 + y2 + 2x + 4y – 3 = 0 is [2008]

Solution:

The given circle is  x2 + y2 + 2x + 4y –3 = 0

Centre (–1, –2)

Let Q (α, β) be the point diametrically opposite to the point P(1, 0),

⇒ α = –3, β = – 4, So, Q is (–3, –4)

QUESTION: 19

The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is 

Solution:

Let the centre of the circle be (h, 2)

∴ Equation of circle is ( x– h)2 + (y – 2)2 = 25 …(1)

Differentiating with respect to x, we get

Substituting in equation (1) we get

⇒ (y – 2)2  (y')2 = 25 –  (y –2)2

QUESTION: 20

If P an d Q are the points of intersection of the circles x2 + y2 + 3x + 7y+ 2p- 5 =0 and x+ y2 + 2x + 2y – p= 0 then there is a circle passing through P, Q and (1, 1) for: [2009]

Solution:

The given circles are S1 ≡ x2 + y2 + 3x + 7y + 2p  – 5 = 0 ....(1)
S2 ≡ x2 + y2 + 2x + 2y – p2 = 0 ....(2)
∴ Equation of common chord PQ is S1 – S2 = 0

⇒ L≡ x+5y+ p2 + 2p -5=0
⇒ Equation of circle passing through P and Q is S1 + λL = 0
⇒ (x2 + y2 + 3x + 7y + 2p  – 5) + λ (x + 5y + p2 +2p – 5) = 0
As it passes through (1, 1), therefore

⇒ (7 + 2p ) + λ (2p + p2 + 1) = 0

 which does not exist for p = – 1

QUESTION: 21

The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if [2010]

Solution:

Circle x2 + y2 - 4 x - 8y - 5=0

Centre = (2, 4), Radius =

If circle is intersecting line 3x - 4 y=m, at two distinct points.
⇒ length of perpendicular from centre to the line < radius

⇒ |10 + m| < 25

⇒ –25 < m  + 10 < 25 ⇒ – 35 < m < 15

QUESTION: 22

The two circles x2 + y2 = ax and x2 + y2 = c2 (c > 0) touch each other if [2011]

Solution:

As centre of one circle is (0, 0) and other circle passes through (0, 0), therefore

Also 

If the two circles touch each other, then they must touch each other internally.

QUESTION: 23

The length of the diameter of the circle which touches the x-axis at the point (1,0) and passes through the point (2,3) is: [2012]

Solution:

Let centre of the circle be (1,h)

Let the circle passes through the point B (2,3)

∴ CA = CB (radius) ⇒ CA2 = CB2
⇒ (1 – 1)2 + (h  – 0)2 = (1 – 2)2 + ( h – 3)2

⇒ h2 = 1 + h2 + 9 – 6h ⇒ 

Thus, diameter is 

QUESTION: 24

The circle passing through (1, –2) and touching the axis of x at (3, 0) also passes through the point [JEE M 2013]

Solution:

Since circle touches x-axis at (3, 0)

∴ The equation of circle be (x – 3)2 + (y – 0)2 + λy = 0

As it passes through (1, –2)

∴ Put x = 1, y = –2

⇒ (1 – 3)2 + (–2)2 + λ(–2) = 0 ⇒ λ = 4

∴ equation of circle is (x – 3)2 + y2 – 8 = 0

Now,  from the options (5, –2) satisfies equation of circle.

QUESTION: 25

Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to [JEE M 2014]

Solution:

Equation of circle C ≡ ( x- 1)2 + (y- 1)2 =1

Radius of T = | y |
T touches C externally therefore,
Distance between the centres = sum of their radii

⇒ (0 – 1)2 + (y –1)2 = (1 + |y|)2
⇒ 1 + y2 + 1 – 2y = 1 + y2 + 2| y |
2 | y | = 1 – 2y

If y > 0 then 2y = 1 – 2y ⇒ y  = 

If y < 0 then –2y = 1 – 2y ⇒ 0 = 1 (not possible)

∴ y = 

QUESTION: 26

Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k (x – 2y + 3) = 0, k ∈ R, is a : [JEE M 2015]

Solution:

Intersection point of 2x – 3y + 4 = 0 and x – 2y + 3 = 0 is (1, 2)

Since, P is the fixed point for given family of lines So, PB = PA
(α – 1)2 + (β – 2)2 = (2 – 1)2 + (3 – 2)2
(α – 1)2 + (β –2)2 = 1 + 1 = 2
(x – 1)2 + (y – 2)2 =
(x – α)2 + (y – β)2 = r2
Therefore, given locus is a circle with centre (1, 2) and radius 

QUESTION: 27

The number of common tangents to the circles x2 + y2 – 4x  – 6x – 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0, is   :[JEE M 2015]

Solution:

x2 + y2 – 4x – 6y – 12 = 0 ...(i)
Centre, c1 = (2, 3) and
Radius, r1 = 5 units x2 + y2 + 6x + 18y + 26 = 0 ...(ii)
Centre, c2 = (–3, –9) and Radius, r2 = 8 units

r1 + r2 = 5 + 8 = 13

∴ C1 C2 = r1 + r2

Therefore there are three common tangents.

QUESTION: 28

The centres of those circles which touch the circle,  x2 + y2 – 8x – 8y – 4 = 0, externally and also touch the x-axis,lie on: [JEE M 2016]

Solution:

For the given circle, centre : (4, 4) radius = 6

(h – 4)2 = 20k + 20
∴ locus of (h, k) is (x – 4)2 = 20(y + 1), which is a parabola.

QUESTION: 29

If one of the diameters of the circle, given by the equation, x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is at (–3, 2), then the radius of S is: [JEE M 2016]

Solution:

Centre of S : O (–3, 2) centre of given circle A(2, –3)

Also AB = 5 (∵ AB = r of the given circle)

⇒ Using pythagoras theorem in ΔOAB

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