Q 11 :

A line segment AB of length λ moves such that the points A and B remain on the periphery of a circle of radius λ. Then the locus of the point that divides the line segment AB in the ratio 2 : 3, is a circle of radius             [2023]

  • 195λ    

     

  • 23λ    

     

  • 197λ    

     

  • 35λ

     

(1)

OAP=60°                          [ OA=OB=AB=λ]

AP=2λ5  (Given)

cos60°=OA2+AP2-OP22·OA·AP

12=λ2+4λ225-OP22λ·2λ5

2λ25=29λ225-OP2OP2=2925λ2-25λ2=1925λ2

Required locus is x2+y2=1925λ2

   Radius=195λ



Q 12 :

Let A be the point (1, 2) and B be any point on the curve x2+y2=16. If the centre of the locus of the point P, which divides the line segment AB in the ratio 3 : 2, is the point  C(α,β), then the length of the line segment AC is          [2023]

  • 255

     

  • 355

     

  • 655

     

  • 455

     

(2)

Since, P(h,k) divides AB in the ratio 3:2.

So, 12cosθ+25=h and 12sinθ+45=k

12cosθ=5h-2                                        ...(i)

and 12sinθ=5k-4                                      ...(ii)

Squaring (i) and (ii) and then adding, we get  

144=(5h-2)2+(5k-4)2

(x-25)2+(y-45)2=14425

Centre (25,45)C(α,β)

AC=(1-25)2+(2-45)2 =925+3625=355



Q 13 :

Let the centre of a circle C be (α,β) and its radius r<8. Let 3x+4y=24 and 3x-4y=32 be two tangents and 4x+3y=1 be a normal to C. Then (α-β+r) is equal to               [2023]

  • 6

     

  • 5

     

  • 7

     

  • 9

     

(3)

After solving 4x+3y=1 and 3x-4y=32, Point A is (4,-5)

Centre (α,β) lies on the line 4x+3y=1

So, 4α+3β=1

   β=1-4α3

Now, distance from Centre to line 3x-4y-32=0 and 3x+4y-24=0 are equal.

So, |3α-4(1-4α3)-3232+42|=|3α+4(1-4α3)-245|

  |3α-4(1-4α3)-325|=|3α+4(1-4α3)-245|

  |9α-4+16α-9615|=|9α+4-16α-7215|

 25α-100=±(-7α-68)

  α=1 and α=283

For α=1, Centre is (1,-1)

and radius is  (4-1)2+(-5+1)2=5 units

For α=283β=-1099, Centre (283,-1099)

  radius≃8.88 (rejected)

Hence, α=1, β=-1, r=5

   α-β+r=7



Q 14 :

The number of common tangents to the circles x2+y2-18x-15y+131=0 and x2+y2-6x-6y-7=0, is          [2023]

  • 2

     

  • 3

     

  • 4

     

  • 1

     

(2)

C1:x2+y2-18x-15y+131=0

C2:x2+y2-6x-6y-7=0

We know that for equation of circle  

x2+y2+2gx+2fy+C=0

Centre is (-g,-f) and radius =g2+f2-C

For C1, Centre (A)=(9,152)

and radius (r1)=81+2254-131=254=52

For C2, Centre (B)=(3,3)

and radius (r2)=9+9+7=5

and AB=(6)2+(92)2=36+814=2254=152=r1+r2

  There are 3 common tangents



Q 15 :

The area enclosed by the closed curve C given by the differential equation dydx+x+ay-2=0, y(1)=0 is 4π. Let P and Q be the points of intersection of the curve C and the y-axis. If normals at P and Q on the curve C intersect the x-axis at points R and S respectively, then the length of the line segment RS is       [2023]

  • 433

     

  • 233

     

  • 2

     

  • 23

     

(1)

dydx+x+ay-2=0, y(1)=0;    dydx=-(x+a)y-2

(y-2)dy=-(x+a)dxy22-2y=-x22-ax+c

x2+y22+ax-2y=cx2+y2+2ax-4y=c1    (c1=2c)

Substituting x=1 and y=0 as y(1)=0

1+2a=c1     x2+y2+2ax-4y=1+2a

This represents the equation of a circle having area 4π

 Radius of circle=2

Now r2=22=(-a)2+(2)2=1+2a

0=(a+1)2  a=-1

  Equation of circle is  x2+y2-2x-4y=-1

 x2-2x+1+y2-4y+4-4=0

 (x-1)2+(y-2)2=4

Now, substitute x=0 to get intersection point of circle with y-axis

(y-2)2=3

y=±3+2      P(0,±3+2)

P(0,±3+2) and Q(0,2-3) are the points.

Now normal at P will pass through centre (1,2)

  Equation of line passing through P and (1,2) is

y-(3+2)=-31(x)

Now, this line will cut x-axis at  R(3+23,0)

Also, equation of line passing through Q(0,2-3)  and (1,2) is y-(2-3)=3x

This will cut the x-axis at  S(3-23,0)

RS=(3+23-3-23)2+(0-0)2=(43)2=43=433



Q 16 :

The locus of the mid points of the chords of the circle C1:(x-4)2+(y-5)2=4 which subtend an angle θi at the centre of the circle C1, is a circle of radius ri. If θ1=π3,θ3=2π3 and r12=r22+r32, then θ2 is equal to            [2023]

  • π2

     

  • π4

     

  • π6

     

  • 0-π4

     

 (1)

If a chord of circle of radius R subtends angle θi at the centre C1, then locus of the end point of this chord in a circle of radius

ri=Rcos(θi2)

Given, r12=r22+r32

cos2θ12=cos2θ22+cos2θ32

cos2π6=cos2θ22+cos2π3

34=cos2θ22+14

cos2θ22=12cosθ22=12θ22=π4θ2=π2



Q 17 :

The points of intersection of the line ax+by=0(ab) and the circle x2+y2-2x=0 are A(α,0) and B(1,β). The image of the circle with AB as a diameter in the line x+y+2=0 is                [2023]

  • x2+y2+3x+3y+4=0

     

  • x2+y2+5x+5y+12=0

     

  • x2+y2+3x+5y+8=0

     

  • x2+y2-5x-5y+12=0

     

(2)

Clearly the line and circle intersect at (0, 0)  

α=0

(1,β) lies on the circle  

1+β2-2×1=0 β2=1 β=±1

Let β=-1

(1, -1) should lie on the line  

a-b=0a=b but ab

Therefore β=1.  

The line and circle intersect at (0, 0) and (1, 1).

Centre of circle with diameter AB is (12,12)

Radius of circle=12, where A(0,0) and B(1,1)

Let (p, q) be the mirror image of (12,12) w.r.t. x+y+2=0

p-1/21=q-1/21=-2(12+12+2)2=-3p=-52, q=-52

(p, q) is centre of required circle whose equation is  

(x+52)2+(y+52)2=12x2+y2+5x+5y+252-12=0

x2+y2+5x+5y+12=0



Q 18 :

Let the tangents at the points A(4,-11) and B(8,-5) on the circle x2+y2-3x+10y-15=0, intersect at the point C. Then the radius of the circle whose centre is C and the line joining A and B is its tangent, is equal to           [2023]

  • 213    

     

  • 2133    

     

  • 13    

     

  • 334

     

(2)

Equation of tangent at point A(4,-11) is

4x-11y-32(x+4)+5(y-11)-15=0

5x-12y=152                                 ...(i)

Equation of tangent at point B(8,-5) is

8x-5y-32(x+8)+5(y-5)-15=0

x=8                                                ...(ii)

Solving (i) and (ii) we get

          x=8 and y=-283

So, point C is (8,-283) and this point is centre of circle whose tangent is line AB.

Equation of line AB is, 3x-2y=34

Perpendicular distance of point C from line AB gives the radius of another circle whose centre is  C(8,-283).

So, |3×8-2(-283)-34|32+(-2)2=2133



Q 19 :

Let y=x+2, 4y=3x+6 and 3y=4x+1 be three tangent lines to the circle (x-h)2+(y-k)2=r2. Then h+k is equal to      [2023]

  • 5

     

  • 5(1+2)

     

  • 6

     

  • 52

     

(1)

We have equation of circle (x-h)2+(y-k)2=r2

We know that the perpendicular distance from centre to the tangent is equal to radius of the circle.

L1:x-y+2=0; L2:3x-4y+6=0; L3:4x-3y+1=0

So, r=h-k+22=3h-4k+69+16=4h-3k+116+9

Consider, 3h-4k+65=4h-3k+15

 3h-4k+6=4h-3k+1h+k=5



Q 20 :

Let a circle C1 be obtained on rolling the circle x2+y2-4x-6y+11=0 upwards 4 units on the tangent T to it at the point (3, 2). Let C2 be the image of C1 in T.  Let A and B be the centers of circles C1 and C2 respectively, and M and N be respectively the feet of perpendiculars drawn from A and B on the x-axis. Then the area of the trapezium AMNB is:         [2023]

  • 2(2+2)

     

  • 2(1+2)

     

  • 4(1+2)

     

  • 3+22

     

(3)

We have, x2+y2-4x-6y+11=0                   ...(i)

Centre = (2, 3); Radius =2

Tangent at (3,2) is, 3x+2y-2(x+3)-3(y+2)+11=0

  x-y=1

On rolling the given circle (i) upwards 4 units on the tangent x-y-1=0, centre of the circle also moves upwards 4 units on T. Let the centre of the new circle C1 be (h,k).

Since, slope of tangent = Slope of line joining two centres of the circle.  

Then the increment in both coordinates will be same.

From figure,

(2+a-2)2+(3+a-3)2=16

a2+a2=16  a2=8  a=22

Hence, the centre of circle C1 is (2+22, 3+22) and radius remains same 2.

Equation of circle C1 is (x-2-22)2+(y-3-22)2=2

The centre of circle C2 is the image of C1 in T, then 

x-2-21=y-3-22-1=-2(2+22-3-22-1)12+(-1)2

  x-2-22=2 and y-3-22=-2

  y=1+22  x=4+22

The centre of circle C2 is  B(4+22,1+22).

Feet of perpendicular from A and B on the x-axis are  M(2+22,0) and N(4+22,0) respectively.

Area of trapezium AMNB =12(3+22+1+22)(4+22-2-22)

=12×(4+42)×2=4+42 sq. units.



Q 21 :

The set of all values of a2 for which the line x+y=0 bisects two distinct chords drawn from a point P(1+a2,1-a2) on the circle 2x2+2y2-(1+a)x-(1-a)y=0,is equal to              [2023]

  • (8,)

     

  • (0,4]

     

  • (4,)

     

  • (2,12]

     

(1)



Q 22 :

Let the point (p,p+1) lie inside the region E={(x,y):3-xy9-x2, 0x3}. If the set of all values of p is the interval (a,b), then b2+b-a2 is equal to _____ .          [2023]



(3)

Given, 3-xy9-x2

3-xyx+y-30  (i)

y9-x2x2+y29

Now, (p,p+1) lies inside the region E.       p+p+1-30

p1 and p2+(p+1)29

2p2+2p-80  p2+p-40

p(-(1+17)2,17-12)

  p(1,17-12)    (as p1)

a=1, b=17-12   b2+b-a2=3



Q 23 :

A circle passing through the point P(α,β) in the first quadrant touches the two coordinate axes at the points A and B. The point P is above the line AB. The point Q on the line segment AB is the foot of perpendicular from P on AB. If PQ is equal to 11 units, then the value of αβ is ________.         [2023]



(121)

Let the equation of the circle be

(x-a)2+(y-a)2=a2,

It passes through P(α,β)

   (α-a)2+(β-a)2=a2

 α2+β2-2aα-2aβ+a2=0                ...(i)

Here, the equation of line AB is x+y=a

Let Q(α',β') be the foot of the perpendicular from P on AB

  α'-α1 =β'-β1=-(α+β-a)2

(PQ)2=(α'-α)2+(β'-β)2=14(α+β-a)2+14(α+β-a)2

(11)2=12(α+β-a)2

242=α2+β2+a2+2αβ-2aα-2aβ

242=2αβ                                                           [From (i)]

αβ=121



Q 24 :

Consider a circle C1:x2+y2-4x-2y=α-5. Let its mirror image in the line y=2x+1 be another circle C2:5x2+5y2-10fx-10gy+36=0. Let r be the radius of C2. Then α+r is equal to _______ .         [2023]



(2)

We have,

       C1:x2+y2-4x-2y-(α-5)=0

   Centre is (2,1), r=4+1+α-5=α

        C2:5x2+5y2-10fx-10gy+36=0

i.e., x2+y2-2fx-2gy+365=0

Centre is (f,g), r=f2+g2-365

The image of (2,1) with respect to 2x-y+1=0 is (f,g),

then f-22=g-1-1=-2(4-1+1)5

f-22=-85 and g-1=85

f-2=-165 and g-1=85

f=2-165=-65 and g=135

So, (f,g)=(-65,135)

and r=3625+16925-365r=1

r=1  

  α+r=1+1=2



Q 25 :

Two circles in the first quadrant of radii r1 and r2 touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line x+y=2. Then r12+r22-r1r2 is equal to _______ .          [2023]



(7)

 

 



Q 26 :

Points P(-3, 2), Q(9, 10) and R(α, 4) lie on a circle C with PR as its diameter. The tangents to C at the points Q and R intersect at the point S. If S lies on the line 2x-ky=1, then k is equal to ______ .          [2023]



(3)

Since PQQR, so mPQ·mQR=-1

10-29+3×10-49-α=-1α=13

Since QOQS, so mOQ·mQS=-1 mQS=-47

Equation of QS: y-10=-47(x-9)4x+7y=106    (i)

Since ORSR

            mOR·mRS=-1

mRS=-8

Equation of RS: y-4=-8(x-13)

8x+y=108    (ii)

Solving (i) and (ii)

         x=252, y=8

Since S(x,y) lies on the line 2x-ky=1

       25-8k=18k=24k=3



Q 27 :

A circle with centre (2, 3) and radius 4 intersects the line x+y=3 at the points P and Q. If the tangents at P and Q intersect at the point S(α,β), then 4α-7β is equal to ______ .                    [2023]



(11)

The equation of circle can be written as,

(x-2)2+(y-3)2=42 x2+y2-4x-6y-3=0

Chord of contact at (α,β) is

αx+βy-2(x+α)-3(y+β)-3=0

(α-2)x+(β-3)y-(2α+3β+3)=0     ...(i)

  But the equation of chord of contact is given:    

        x+y-3=0     ...(ii)

Comparing equations (i) and (ii):

α-21= β-31=-(2α+3β+3-3)

On solving, α=-6, β=-5

So, 4α-7β =4(-6)-7(-5) =11



Q 28 :

Let P(a1,b1) and Q(a2,b2) be two distinct points on a circle with center C(2,3). Let O be the origin and OC be perpendicular to both CP and CQ. If the area of the triangle OCP is 352, then a12+a22+b12+b22 is equal to ________ .         [2023]



(24)

Area of OCP=352

12×PC×OC=352

12×PC×5=352

PC=7

Now, OQ2=5+7=12

            OP2=5+7=12

             a12+b12=OP2 and a22+b22=OQ2

   a12+a22+b12+b22 =OP2+OQ2 =12+12=24



Q 29 :

Let PQ and MN be two straight lines touching the circle x2+y2-4x-6y-3=0 at the points A and B respectively. Let O be the centre of the circle and angle AOB=π3 Then the locus of the point of intersection of the lines PQ and MN is:         [2026]

  • 3(x2+y2)-12x-18y-25=0

     

  • 3(x2+y2)-18x-12y+25=0

     

  • x2+y2-12x-18y-25=0

     

  • x2+y2-18x-12y-25=0

     

(1)

Given Circle

x2+y2-4x-6y-3=0

C(2,3) and r=4

cos30°=rOR=4OR

OR=83

Now

OR2=(h-2)2+(k-3)2

3(x2+y2)-12x-18y-25=0



Q 30 :

Let a circle of radius 4 pass through the origin O, the points A(-3a,0) and B(0,-2b), where a and b are real parameters and ab0. Then the locus of the centroid of OAB is a circle of radius               [2026]

  • 73

     

  • 83

     

  • 53

     

  • 113

     

(2)

AB=8

3a2+2b2=64

Centroid G(h,k)

h=-3a3,  k=-2b3

a=-3h,  b=-32k

9h2+9k2=64

x2+y2=649

r=83



Q 31 :

Let the set of all values of r, for which the circles (x+1)2+(y+4)2=r2  and  x2+y2-4x-2y-4=0 intersect at two distinct points be the interval (α,β). Then αβ is equal to              [2026]

  • 24

     

  • 21

     

  • 20

     

  • 25

     

(4)

(x-2)2+(y-1)2=32 & (x+1)2+(y+4)2=r2

|r1-r2|<c1c2<r1+r2

|r-3|<(2+1)2+(1+4)2<r+3

|r-3|<34 & r+3>34

-34<r-3<34 & r>34-3

i.e. r=(3-34,3+34)(34-3,)

i.e. r(34-3,34+3)

 αβ=(34-3)(34+3)

=34-9

=25



Q 32 :

Let the circle x2+y2=4 intersect x-axis at the points A(a,0), a>0 and B(b,0).  Let P(2cosα,2sinα) 0<α<π2 and Q(2cosβ,2sinβ) be two points such that (α-β)=π2.  Then the point of intersection of AQ and BP lies on :   [2026]

  • x2+y24x4=0

     

  • x2+y24y4=0

     

  • x2+y24x4y-4=0

     

  • x2+y24x4y=0

     

(2)

Let point of intersection R(h,k)

mBR=mBPkh+2=2sinα2cosα+2kh+2=tanα2

mAR=mAQkh-2=2sinβ2cosβ-2=sinβcosβ-1=-cotβ2

α2-β2=π4

tan(α2-β2)=tanπ4=1

tanα2-tanβ21+tanα2tanβ2=1

kh+2+h-2k1+(kh+2)(2-hk)=1

k2+h2-4k(h+2)4h+2=1

h2+k2-44k=1

x2+y2-4y-4=0



Q 33 :

Let (h,k) lie on the circle C:x2+y2=4 and the point (2h+1,3k+2) lie on an ellipse with eccentricity e. Then the value of 5e2 is equal to _________      [2026]



(9)

Let P(2cosθ,2sinθ)

 coordinates of Q=(4cosθ+1,6sinθ+3)

 locus of Q is (x-14)2+(y-36)2=1

 e2=1-1636=59

 5e2=9



Q 34 :

If P is a point on the circle x2+y2=4,, Q is a point on the straight line 5x+y+2=0 and xy+1=0 is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P is __________.    [2026]



(2)

Mid point of PQ lies on x-y+1=0

2cosθ+α2-2sinθ-5α-22+1=0

2cosθ+α-2sinθ+5α+2+2=0

cosθ-sinθ+3α+2=0          ...(1)

 Slope of PQ is -1

2sinθ+5α+22cosθ-α=-1

2sinθ+5α+2=-2cosθ+α

sinθ+cosθ+2α+1=0          ...(2)

Eliminate α from (1) and (2)

cosθ+5sinθ=1,  θ[0,2π]

5×2sinθ2cosθ2=2sin2θ2

 sinθ2=0cosθ=1

or

sinθ2=5cosθ=-1213

Sum of all possible values of abscissa of point P is

=2×1+2(-1213)=213

 13 times sum of all possible values of abscissa of point P is=2



Q 35 :

Let y=x be the equation of a chord of the circle C1 (in the closed half-plane x0) of diameter 10 passing through the origin. Let C2 be another circle described on the given chord as its diameter. If the equation of the chord of the circle C2, which passes through the point (2,3) and is farthest from the center of C2, is x+ay+b=0, then a−b is equal to     [2026]

  • 10

     

  • -2

     

  • 6

     

  • -6

     

(2)

Equation of circle C2 is 

x2+y2-5x-5y=0

Its centre is (52,52)

mAB=-1

 Slope of required chord =1

 Equation of required chord is x-y+1=0

 a=-1, b=2

 a-b=-2