JEE Main Previous year questions (2021-22): Circle - Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Q.1. Let ABC be the triangle with AB = 1, AC = 3 and ∠BAC = π/2. If a circle of radius r > 0 touches the sides AB, AC and also touches internally the circumcircle of the triangle ABC, then the value of r is __________.         (JEE Advanced 2022)

Ans. Between 0.82 and 0.86

Here ABC is a right angle triangle. BC is the Hypotenuse of the triangle.

We know, diameter of circumcircle of a right angle triangle is equal to the Hypotenuse of the triangle also midpoint of Hypotenuse is the center of circle.

∴ BC = Diameter of the circle

Here B = (0, 1) and C = (3,0)

Radius of circumcircle (R) = √10/2

∴ Center of circle (M) =

Center of circle which touches line AB and AC = (r, r)

Now distance between center of two circles,

⇒ r2 − 4r + √10r = 0

⇒ r(r − 4 + √10) = 0

⇒ r = 0 or r = r − √10

∴ r = 4 − √10 [as r ≠ 0]

= 0.837
≃ 0.842

Q.2. Let G be a circle of radius R > 0. Let G₁, G₂,…, G_n be n circles of equal radius r > 0. Suppose each of the n circles G₁, G₂,…, G_n touches the circle G externally. Also, for i = 1, 2,…, n − 1, the circle G_i touches G_i+1 externally, and G_n touches G₁ externally. Then, which of the following statements is/are TRUE?         (JEE Advanced 2022)
(a) If n = 4, then (√2 − 1)r < R
(b) If n = 5, then r < R
(c) If n = 8, then (√2 − 1)r < R
(d) If n = 12, then √2(√3 + 1)r > R

Ans. c, d

Here if we add center of circles G₁, G₂, G₃ ....... G_n, then we get a polygon of n sides.

From figure you can see one side of polygon makes angle θ with the center.

∴ n sides make angle = nθ

We know, nθ = 2π

⇒ θ = 2π/n

Here triangle OMN is an isosceles triangle. Line joining of point O and midpoint O of MN (point A) is perpendicular to line MN and perpendicular bisector of angle θ.

∴ ∠MOA = θ/2 = π/n

From right angle triangle OMA,

Option A:

When n = 4,

∴ Option (A) is wrong.

Option B:

When n = 5,

We know, in 0 to π/2, cosecθ is decreasing.

∴ cosec45^∘ < cosec36^∘ < cosec30^∘

∴ R < r, when n = 5

∴ Option (B) is wrong.

Option C:

When n = 8,

cosecθ is decreasing function in 0 to π/2.

∴ Option (C) is correct.

Option D:

When n = 12, then

Option (D) is correct.

Q.3. Let Let S = {(x, y) ∈ N × N : 9(x − 3)² + 16(y − 4)² ≤ 144} and T = {(x, y) ∈ R × R : (x − 7)² + (y − 4)² ≤ 36}. Then n(S ∩ T) is equal to __________.         (JEE Main 2022)

Ans. 27

Q.4. Let AB be a chord of length 12 of the circle (x − 2)² + (y + 1)² = 169/4. If tangents drawn to the circle at points A and B intersect at the point P, then five times the distance of point P from chord AB is equal to __________.         (JEE Main 2022)

Ans. 72
Here AM = BM = 6


sin⁡θ = 12/13
In △PAO:
PO/OA = sec⁡θ

Q.5. Let the mirror image of a circle c₁ : x² + y² − 2x − 6y + α = 0 in line y = x + 1 be c₂ : 5x² + 5y² + 10gx + 10fy + 38 = 0. If r is the radius of circle c₂, then α + 6r² is equal to ________.         (JEE Main 2022)

Ans. 12
c₁ : x² + y² − 2x − 6y + α = 0

Then centre = (1, 3) and radius (r) =

Image of (1, 3) w.r.t. line x − y + 1 = 0 is (2, 2)

c₂ : 5x² + 5y² + 10gx + 10fy + 38 = 0

Then (−g, −f) = (2, 2)

∴ g = f = −2 .......... (i)

Q.6. If the circles x² + y² + 6x + 8y + 16 = 0 and x² + y² + 2(3 − √3)x + 2(4 − √6)y = k + 6√3 + 8√6, k > 0, touch internally at the point P(α, β), then (α + √3)² + (β + √6)² is equal to ________________.         (JEE Main 2022)

Ans. 25

The circle x² + y² + 6x + 8y + 16 = 0 has centre (−3,−4) and radius 3 units.

The circle x² + y² + 2(3 − √3)x + 2(4 − √6)y = k + 6√3 + 8√6, k > 0 has centre
 These two circles touch internally hence 

Here, k = 2 is only possible (∵ k > 0)

Equation of common tangent to two circles is 2√3x + 2√6y + 16 + 6√3 + 8√6 + k = 0

∵ k = 2 then equation is

x + √2y + 3 + 4√2 + 3√3 = 0 ...... (i)

∵ (α, β) are foot of perpendicular from (−3, −4)

To line (i) then

⇒ (α + √3)² = 9 and (β + √6)² = 16
∴ (α + √3)² + (β + √6)² = 25

Q.7. If one of the diameters of the circle x² + y² − 2√2x − 6√2y + 14 = 0 is a chord of the circle (x−2√2)² + (y − 2√2)² = r², then the value of r² is equal to ____________.         (JEE Main 2022)

Ans. 10

Q.8. Let the lines y + 2x = √11 + 7√7 and 2y + x = 2√11 + 6√7 be normal to a circle C : (x − h)² + (y − k)² = r². If the line √11y − 3x =is tangent to the circle C, then the value of (5h − 8k)² + 5r² is equal to __________.         (JEE Main 2022)

Ans. 816

L₁: y + 2x = √11 + 7√7

L₂: 2y + x = 2√11 + 6√7

Point of intersection of these two lines is centre of circle i.e.

⊥^r from centre to lineis radius of circle

So (5h − 8K)² + 5r²

= 64 × 11 + 112 = 816

Q.9. Let a circle C of radius 5 lie below the x-axis. The line L₁ : 4x + 3y + 2 = 0 passes through the centre P of the circle C and intersects the line L₂ = 3x − 4y − 11 = 0 at Q. The line L₂ touches C at the point Q. Then the distance of P from the line 5x − 12y + 51 = 0 is ______________.         (JEE Main 2022)

Ans. 11
L₁: 4x + 3y + 2 = 0

L₂: 3x − 4y − 11 = 0

Since circle C touches the line L₂ at Q intersection point Q of L₁ and L₂, is (1, −2)

∵ P lies of L₁

⇒ (x − 1)² = 9

⇒ x = 4, −2

∵ Circle lies below the x-axis

∴ y = −6

P(4, −6)

Now distance of P from 5x − 12y + 51 = 0

Q.10. A rectangle R with end points of one of its sides as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is 2x − y + 4 = 0, then the area of R is ____________.         (JEE Main 2022)

Ans. 16

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side 

Q.11. Let the abscissae of the two points P and Q be the roots of 2x² − rx + p = 0 and the ordinates of P and Q be the roots of x² − sx − q = 0. If the equation of the circle described on PQ as diameter is 2(x² + y²) − 11x − 14y − 22 = 0, then 2r + s − 2q+p is equal to __________.         (JEE Main 2022)

Ans. 7
Let P(x₁, y₁) & Q(x₂, y₂)

∴ Roots of 2x² − rx + p = 0 are x₁, x₂

and roots of x² − sx − q = 0 are y₁, y₂.

∴ Equation of circle ≡ (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0

⇒ x² − (x₁ + x₂)x + x₁x₂ + y² − (y₁ + y₂)y + y₁y₂ = 0

⇒ 2x² + 2y² − rx + 2sy + p − 2q = 0

Compare with 2x² + 2y² − 11x − 14y − 22 = 0

We get r = 11, s = 7, p − 2q = −22
⇒ 2r + s + p − 2q = 22 + 7 − 22 = 72

Q.12. Let a circle C : (x − h)² + (y − k)² = r², k > 0, touch the x-axis at (1, 0). If the line x + y = 0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h + k + r is equal to ___________.         (JEE Main 2022)

Ans. 7

Here, OM² = OP² − PM²

∴ r² − 2r − 3 = 0

∴ r = 3

∴ Equation of circle is

(x − 1)² + (y − 3)² = 3²

∴ h = 1, k = 3, r = 3

∴ h + k + r = 7

Q.13. Let the tangents at two points A and B on the circle x² + y2 − 4x + 3 = 0 meet at origin O(0,0). Then the area of the triangle OAB is:         (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans. b
x² + y2 − 4x + 3 = 0

⇒ (x − 2)² + y² = 1


Also, AO = BO

Q.14. The foot of the perpendicular from a point on the circle x² + y² = 1, z = 0 to the plane 2x + 3y + z = 6 lies on which one of the following curves?         (JEE Main 2022)
(a) (6x + 5y − 12)² + 4(3x + 7y − 8)² = 1, z = 6 − 2x − 3y
(b) (5x + 6y − 12)² + 4(3x + 5y − 9)² = 1, z = 6 − 2x − 3y
(c) (6x + 5y − 14)² + 9(3x + 5y − 7)² = 1, z = 6 − 2x − 3y
(d) (5x + 6y − 14)² + 9(3x + 7y − 8)² = 1, z = 6 − 2x − 3y

Ans. b
Any point on x² + y² = 1, z = 0 is p(cos⁡θ, sin⁡θ, 0)

If foot of perpendicular of p on the plane 2x + 3y + z = 6 is (h, k, l) then

h = 2r + cos⁡θ, k = 3r + sin⁡θ, l = r

Hence, h − 2l = cos⁡θ and k − 3l = sin⁡θ

Hence, (h − 2l)² + (k − 3l)² = 1

When l = 6 − 2h − 3k

Hence required locus is

(x − 2(6 − 2x − 3y))² + (y − 3(6 − 2x −3y))² = 1

⇒ (5x + 6y − 12)² + 4(3x + 5y − 9)² = 1, z = 6 − 2x − 3y

Q.15. Let C be the centre of the circle x² + y² − x + 2y = 11/4 and P be a point on the circle. A line passes through the point C, makes an angle of π/4 with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit 2 ) is:         (JEE Main 2022)
(a) 2
(b) 2√2
(c) 8sin⁡(π/8)
(d) 8cos⁡(π/8)

Ans. b

QR = 2r = 4

= 2√2 sq. units

Q.16. For t ∈ (0, 2π), if ABC is an equilateral triangle with vertices A(sin⁡t, −cos⁡t), B(cost, sin⁡t) and C(a, b) such that its orthocentre lies on a circle with centre (1, 1/3), then (a² − b²) is equal to:        (JEE Main 2022)
(a) 2
(a) 8/3
(b) 8
(c) 77/9
(d) 80/9

Ans. b
Let P(h, k) be the orthocentre of ΔABC 

Then

(orthocentre coincide with centroid)

∴ (3h − a)² + (3k − b)² = 2

∵ orthocentre lies on circle with centre (1, 1/3)

∴ a = 3, b = 1

∴ a² − b² = 8

Q.17. A circle C₁ passes through the origin O and has diameter 4 on the positive x-axis. The line y = 2x gives a chord OA of circle C₁. Let C₂ be the circle with OA as a diameter. If the tangent to C₂ at the point A meets the x-axis at P and y-axis at Q, then QA : AP is equal to:       (JEE Main 2022)
(a) 1 : 4
(b) 1 : 5
(c) 2 : 5
(d) 1 : 3

Ans. a

Equation of C₁

x² + y² − 4x = 0

Intersection with

y = 2x

x² + 4x² − 4x = 0

5x² − 4x = 0

x + 2y = 4 ⇒ P : (4, 0), Q : (0, 2)
QA : AP = 1 : 4

Q.18. If the circle x² + y² − 2gx + 6y − 19c = 0, g, c ∈ R passes through the point (6, 1) and its centre lies on the line x − 2cy = 8, then the length of intercept made by the circle on x-axis is       (JEE Main 2022)
(a) √11
(b) 4
(c) 3
(d) 2√23

Ans. d

Circle: x² + y² − 2gx + 6y − 19c = 0

It passes through h(6, 1)

⇒ 36 + 1 − 12g + 6 − 19c = 0

= 12g + 19c = 43 ..... (1)

Line x − 2cy = 8 passes through centre

⇒ g + 6c = 8 ...... (2)

From (1) & (2)

g = 2, c = 1

C: x² + y² − 4x + 6y − 19 = 0

Q.19. Let the abscissae of the two points P and Q on a circle be the roots of x² − 4x − 6 = 0 and the ordinates of P and Q be the roots of y² + 2y − 7 = 0. If PQ is a diameter of the circle x² + y² + 2ax + 2by + c = 0, then the value of (a + b − c) is _____________.       (JEE Main 2022)
(a) 12
(b) 13
(c) 14
(d) 16

Ans. a
Abscissae of PQ are roots of x² − 4x − 6 = 0

Ordinates of PQ are roots of y² + 2y − 7 = 0

and PQ is diameter

⇒ Equation of circle is

x² + y² − 4x + 2y − 13 = 0

But, given x² + y² + 2ax + 2by + c = 0

By comparison a = −2, b = 1, c = −13
⇒ a + b − c = −2 + 1 + 13 = 12

Q.20. Consider three circles:
C₁: x² + y² = r²
C₂: (x − 1)² + (y − 1)² = r²
C₃: (x − 2)² + (y − 1)² = r²
If a line L : y = mx + c be a common tangent to C₁, C₂ and C₃ such that C₁ and C₃ lie on one side of line L while C₂ lies on other side, then the value of 20(r² + c) is equal to:       (JEE Main 2022)
(a) 23
(b) 15
(c) 12
(d) 6

Ans. d

C₁: x² + y² = r² ; center = (0, 0) and radius = r

C₂: (x − 1)² + (y − 1)² = r² ; center = (1, 1) and radius = r

C₃: (x − 2)² + (y − 1)² = r² ; center = (2, 1) and radius = r

Distance of y = mx + c line from center (0, 0) is,

Distance of y = mx + c line from center (1, 1) is,

Distance of y = mx + c line from center (2, 1) is,

From (1) and (2), we get

⇒ m − 1 + c = ±c ..... (4)

taking positive sign,

m − 1 + c = c

⇒ m − 1 = 0

⇒ m = 1

From (2) and (3), we get

⇒ (m − 1 + c) = ±(2m − 1 + c) ...... (5)

taking positive sign,

m − 1 + c = 2m − 1 + c

⇒ m = 0

By taking positive sign we get two different value of m so it is not acceptable.

From equation (4), taking negative sign,

m − 1 + c = −c

⇒ m + 2c − 1 = 0 ..... (6)

From equation (5), taking negative sign

m − 1 + c = −(2m − 1 + c)

⇒ 3m + 2c − 2 = 0 ..... (7)

Solving equation (6) and (7), we get

3m + 1 − m − 2 = 0

⇒ 2m = 1

⇒ m = 1/2

⇒ c = 1/4

Putting value of m = 1/2 and c = 1/4 in equation (1), we get

∴ 20(r² + c)

= 6

Q.21. Let a triangle ABC be inscribed in the circle x² − √2(x + y) + y² = 0 such that ∠BAC = π/2. If the length of side AB is √2, then the area of the ΔABC is equal to:       (JEE Main 2022)
(a) 1
(b) (√6 + √3)/2
(c) (3 + √3)/4
(d) (√6 + 2√3)/4

Ans. a
Note:

For equation of circle x² + y² + 2gx + 2fy + c = 0, center is (−g, −f) and radius

Given,

equation of circle is

As AB and AC makes an angle 90∘ then line BC passes through the center of circle and BC is the diameter of the circle.

∴ Length of BC = 2r = 2 × 1 = 2

∴ AC² = BC² − AB²

= 2² − (√2)²

= 2

⇒ AC = 2

∴ Area of right angle triangle ABC

= 1 square unit.

Q.22. Let the tangent to the circle C₁ : x² + y² = 2 at the point M(−1, 1) intersect the circle C₂ : (x − 3)² + (y − 2)² = 5, at two distinct points A and B. If the tangents to C₂ at the points A and B intersect at N, then the area of the triangle ANB is equal to:       (JEE Main 2022)
(a) 1/2
(b) 2/3
(c) 1/6
(d) 5/3

Ans. c

Equation of tangent at point M is

T = 0

⇒ xx₁ + yy₁ = 2

⇒ −x + y = 2

⇒ y = x + 2

Putting this value to equation of circle C₂,

(x − 3)² + (y − 2)² = 5

⇒ (x − 3)² + x² = 5

⇒ x² − 6x + 9 + x² = 5

⇒ 2x² − 6x + 4 = 0

⇒ x² − 3x + 2 = 0

⇒ (x − 2)(x − 1) = 0

⇒ x = 1, 2

when x = 1, y = 3

and when x = 2, y = 4

∴ Point A(1, 3) and B(2, 4)

Now, equation of tangent at A(1, 3) on circle (x − 3)² + (y − 2)² = 5 or x² + y² − 6x − 4y + 8 = 0 is

T = 0

xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + C = 0

⇒ x + 3y − 3(x + 1) − 2(y + 3) + 8 = 0

⇒ x + 3y − 3x − 3 − 2y − 6 + 8 = 0

⇒ −2x + y − 1 = 0

⇒ 2x − y + 1 = 0 ....... (1)

Similarly tangent at B(2, 4) is

2x + 4y − 3(x + 2) − 2(y + 4) + 8 = 0

⇒ 2x + 4y − 3x − 6 − 2y − 8 + 8 = 0

⇒ −x + 2y − 6 = 0

⇒ x − 2y + 6 = 0 ...... (2)

Solving equation (1) and (2), we get

x − 2(2x + 1) + 6 = 0

⇒ x − 4x − 2 + 6 = 0

⇒ −3x + 4 = 0

⇒ x = 4/3

Now area of the triangle ANB

Q.23. If the tangents drawn at the points O(0, 0) and P(1 + √5, 2) on the circle x² + y² − 2x − 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to       (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans. c

Q.24. The set of values of k, for which the circle C: 4x² + 4y² − 12x + 8y + k = 0 lies inside the fourth quadrant and the point (1,) lies on or inside the circle C, is       (JEE Main 2022) 
(a) an empty set 
(b) (6, 65/9] 
(c) [80/9, 10) 
(d) (9, 92/9]

Ans. d
C: 4x² + 4y² − 12x + 8y + k = 0
∵ (1, ) lies on or inside the C
 

Now, circle lies in 4^th quadrant centre ≡ (3/2, −1)

Q.25. Let C be a circle passing through the points A(2, −1) and B(3, 4). The line segment AB s not a diameter of C. If r is the radius of C and its centre lies on the circle (x − 5)² + (y − 1)² = 13/2, then r² is equal to:       (JEE Main 2022)
(a) 32
(b) 65/2
(c) 61/2
(d) 30

Ans. b
Equation of perpendicular bisector of AB is

Solving it with equation of given circle, 

But x ≠ 5/2 because AB is not the diameter.

So, centre will be (15/2, 1/2)

= 65/2

Q.26. A circle touches both the y-axis and the line x + y = 0. Then the locus of its center is:       (JEE Main 2022)
(a) y = √2x
(b) x = √2y
(c) y² − x² = 2xy
(d) x² − y² = 2xy

Ans. d
Let the centre be (h, k)

⇒ 2h² = h² + k² + 2hk

Locus will be x² − y² = 2xy

Q.27. Let a, b and c be the length of sides of a triangle ABC such thatIf r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of R/r is equal to:       (JEE Main 2022)
(a) 5/2
(b) 2
(c) 3/2
(d) 1

Ans. a

a + b = 7λ

b + c = 8λ

c + a = 9λ

a + b + c = 12λ

∴ a = 4λ, b = 3λ, c = 5λ

Q.28. Consider the region R = {(x, y) ∈ R × R : x ≥ 0 and y² ≤ 4 − x}. Let F be the family of all circles that are contained in R and have centers on the x-axis. Let C be the circle that has largest radius among the circles in F. Let (α, β) be a point where the circle C meets the curve y² = 4 − x.
The value of α is ___________.       (JEE Advanced 2021)

Ans. 2.00
Given, x ≥ 0, y² ≤ 4 − x
Let equation of circle be
(x − h)² + y² = h² .... (i)

Solving Eq. (i) with y² = 4 − x, we get
x² − 2hx + 4 − x = 0
⇒ x² − x(2h + 1) + 4 = 0 .... (ii)
For touching/tangency, Discriminant (D) = 0
i.e. (2h + 1)² = 16 ⇒ 2h + 1 = ± 4
⇒ 2h = ± 4 − 1
⇒ h = 3/2, h = −5/2 (Rejected) because part of circle lies outside R. So, h = 3/2 = radius of circle (c).
Putting h = 3/2 in Eq. (ii),
x² − 4x + 4 = 0 ⇒ (x − 2)² = 0 ⇒ x = 2
So, α = 2 

Q.29. Consider the region R = {(x, y) ∈ R × R : x ≥ 0 and y² ≤ 4 − x}. Let F be the family of all circles that are contained in R and have centers on the x-axis. Let C be the circle that has largest radius among the circles in F. Let (α, β) be a point where the circle C meets the curve y² = 4 − x.
The radius of the circle C is ___________.        (JEE Advanced 2021)

Ans. 1.50
Given, x ≥ 0, y² ≤ 4 − x
Let equation of circle be
(x − h)² + y² = h² .... (i)

Solving Eq. (i) with y² = 4 − x, we get
x² − 2hx + 4 − x = 0
⇒ x² − x(2h + 1) + 4 = 0 .... (ii)
For touching/tangency, Discriminant (D) = 0
i.e. (2h + 1)² = 16 ⇒ 2h + 1 = ± 4
⇒ 2h = ± 4 − 1
⇒ h = 3/2, h = −5/2 (Rejected) because part of circle lies outside R. So, h = 3/2 = radius of circle (c).

Q.30. Let M = {(x, y) ∈ R × R : x² + y² ≤ r²}, where r > 0. Consider the geometric progression , n = 1, 2, 3, ...... . Let S₀ = 0 and for n ≥ 1, let S_n denote the sum of the first n terms of this progression. For n ≥ 1, let C_n denote the circle with center (S_n−1, 0) and radius a_n, and D_n denote the circle with center (S_n−1, S_n−1) and radius a_n.
Consider M withThe number of all those circles D_n that are inside M is       (JEE Advanced 2021)
(a) 198
(b) 199
(c) 200
(d) 201

Ans. b



∴ n ≤ 199
So, number of circles = 199 

Q.31. Let M = {(x, y) ∈ R × R : x² + y² ≤ r²}, where r > 0. Consider the geometric progression , n = 1, 2, 3, ...... . Let S₀ = 0 and for n ≥ 1, let S_n denote the sum of the first n terms of this progression. For n ≥ 1, let C_n denote the circle with center (S_n−1, 0) and radius a_n, and D_n denote the circle with center (S_n−1, S_n−1) and radius a_n.
Consider M with r=1025/513. Let k be the number of all those circles C_n that are inside M. Let l be the maximum possible number of circles among these k circles such that no two circles intersect. Then       (JEE Advanced 2021)
(a) k + 2l = 22 
(b) 2k + l = 26 
(c) 2k + 3l = 34 
(d) 3k + 2l = 40

Ans. d
 
For circle C_n to be inside M. 


 Number of circles inside be 10 = k. Clearly, alternate circle do not intersect each other i.e. C₁, C₃, C₅, C₇, C₉ do not intersect each other as well as C₂, C₄, C₆, C₈ and C₁₀ do not intersect each other.
Hence, maximum 5 set of circles do not intersect each other.
∴ l = 5
So, 3k + 2l = 40 

Q.32. Consider a triangle Δ whose two sides lie on the x-axis and the line x + y + 1 = 0. If the orthocenter of Δ is (1, 1), then the equation of the circle passing through the vertices of the triangle Δ is       (JEE Advanced 2021)
(a) x² + y² − 3x + y = 0
(b) x² + y² + x + 3y = 0
(c) x² + y² + 2y − 1 = 0
(d) x² + y² + x + y = 0

Ans. b
Equation of circle passing through C(0, 0) is
x² + y² + 2gx + 2fy = 0 ..... (i)
Since Eq. (i), also passes through (−1, 0) and (1, −2).
 
Then, 1 − 2g = 0 ⇒ g = 1 / 2
and 5 + 1 − 4f = 0 ⇒ f = 3 / 2
∴ Equation of circumcircle is
i.e. x² + y² + x + 3y = 0

Q.33. Let Z be the set of all integers,
A = {(x, y) ∈ Z × Z: (x − 2)² + y² ≤ 4}
B = {(x, y) ∈ Z × Z : x² + y² ≤ 4}
C = {(x, y) ∈ Z × Z : (x − 2)² + (y − 2)² ≤ 4}
If the total number of relation from A ∩ B to A ∩ C is 2^p, then the value of p is:       (JEE Main 2021)
(a) 16
(b) 25
(c) 49
(d) 9

Ans. b

(x − 2)² + y² ≤ 4
x² + y² ≤ 4
No. of points common in C₁ & C₂ is 5.
(0, 0), (1, 0), (2, 0), (1, 1), (1, −1)
Similarly in C₂ & C₃ is 5.
No. of relations = 2^5×5 = 2²⁵.

Q.34. A circle C touches the line x = 2y at the point (2, 1) and intersects the circle
C₁ : x² + y² + 2y − 5 = 0 at two points P and Q such that PQ is a diameter of C₁. Then the diameter of C is:        (JEE Main 2021)
(a) 7√5
(b) 15
(c) √285
(d) 4√15

Ans. a
(x − 2)² + (y − 1)² + λ(x − 2y) = 0
C : x² + y² + x(λ − 4) + y(−2 −2λ) + 5 = 0
C₁ : x² + y² + 2y − 5 = 0
S₁ − S₂ = 0 (Equation of PQ)
(λ − 4)x − (2λ + 4)y + 10 = 0 Passes through (0, −1)
⇒ λ = −7
C : x² + y² − 11x + 12y + 5 = 0

Diameter = 7√5

Q.35. If a line along a chord of the circle 4x² + 4y² + 120x + 675 = 0, passes through the point (−30, 0) and is tangent to the parabola y² = 30x, then the length of this chord is:         (JEE Main 2021)
(a) 5
(b) 7
(c) 5√3
(d) 3√5

Ans. d
Equation of tangent to y² = 30 x 

Q.36. Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept 6√5 on the x-axis. Then the radius of the circle C is equal to:         (JEE Main 2021)
(a) √53
(b) 9
(c) 8
(d) √82

Ans. b

Q.37. Let the circle S : 36x² + 36y² − 108x + 120y + C = 0 be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, x − 2y = 4 and 2x − y = 5 lies inside the circle S, then:          (JEE Main 2021)
(a)
(b) 100 < C < 165
(c) 81 < C < 156
(d) 100 < C < 156

Ans. d
S : 36x² + 36y² − 108x + 120y + C = 0 




Now, 


⇒ C > 100 ...... (1)
Now, point of intersection of x − 2y = 4 and 2x − y = 5 is (2, −1), which lies inside the circle S.
∴ S(2, −1) < 0

 
C < 156 ..... (2)
From (1) & (2)
100 < C < 156 Ans. 

Q.38. Let r₁ and r₂ be the radii of the largest and smallest circles, respectively, which pass through the point (−4, 1) and having their centres on the circumference of the circle x² + y² + 2x + 4y − 4 = 0. Ifthen a + b is equal to:          (JEE Main 2021) 
(a) 3
(b) 11
(c) 5
(d) 7

Ans. c

Centre of smallest circle is A
Centre of largest circle is B
r₂ = |CP − CA| = 3√2 − 3
r₁ = CP − CB = 3√2 + 3

a = 3, b = 2

Q.39. Let S₁ : x² + y² = 9 and S₂ : (x − 2)² + y² = 1. Then the locus of center of a variable circle S which touches S₁ internally and S₂ externally always passes through the points:          (JEE Main 2021) 
(a)
(b) (1, ± 2) 
(c) (2, ±3/2) 
(d) (0, ±√3)

Ans. c
S₁ : x² + y² = 9 ; C₁ (0, 0), r₁ = 3
S₂ : (x − 2)² + y² = 1 ; C₂ (2, 0), r₂ = 1
Image
Let the variable circle S and its radius is r units.
Here S and S₁ touches internally
∴ Distance between center,
S + S₁ = PC₁ = 3 − r
Here S and S₂ touches externally
∴ Distance between center,
S + S₂ = PC₂ = 1 + r
∴ PC₁ + PC₂ = 4 > C₁ C₂
So locus is ellipse whose focii are C₁ & C₂ and major axis is 2a = 4 and 2ae = C₁C₂ = 2
⇒ e = 1/2

Centre of ellipse is midpoint of C₁ & C₂ is (1, 0)
Equation of ellipse is
Now by cross checking the option (2, ±3/2) satisfied it.

Q.40. Let A = {(x, y) ∈ R × R|2x² + 2y² − 2x − 2y = 1}, 
B = {(x, y) ∈ R × R|4x² + 4y² − 16y + 7 = 0} and 
C = {(x, y) ∈ R × R|x² + y² − 4x − 2y + 5 ≤ r²}.
Then the minimum value of |r| such that A ∪ B ⊆ C is equal to          (JEE Main 2021)  
(a)
(b)
(c)
(d) 1 + √5

Ans. c




S₃ = x² + y² − 4x − 2y + 5 − r² = 0
C₃ (2, 1) 

Q.41. Two tangents are drawn from the point P(−1, 1) to the circle x² + y² − 2x − 6y + 6 = 0. If these tangents touch the circle at points A and B, and if D is a point on the circle such that length of the segments AB and AD are equal, then the area of the triangle ABD is equal to:           (JEE Main 2021)  
(a) 2
(b)
(c) 4
(d)

Ans. c

Q.42. For the four circles M, N, O and P, following four equations are given:
Circle M : x² + y² = 1
Circle N : x² + y² − 2x = 0
Circle O : x² + y² − 2x − 2y + 1 = 0
Circle P : x² + y² − 2y = 0
If the centre of circle M is joined with centre of the circle N, further center of circle N is joined with centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a:            (JEE Main 2021)  
(a) Rhombus
(b) Square
(c) Rectangle
(d) Parallelogram

Ans. b
C_M = (0, 0)
C_N = (1, 0)
C_O = (1, 1)
C_P = (0, 1)

Q.43. Choose the correct statement about two circles whose equations are given below:
x² + y² − 10x − 10y + 41 = 0
x² + y² − 22x − 10y + 137 = 0             (JEE Main 2021)  
(a) circles have same centre
(b) circles have no meeting point
(c) circles have only one meeting point
(d) circles have two meeting points

Ans. c
Let S₁: x² + y² − 10x − 10y + 41 = 0
⇒ (x − 5)² + (y − 5)² = 9
Centre (C₁) = (5, 5)
Radius r₁ = 3
S₂: x² + y² − 22x − 10y + 137 = 0
⇒ (x − 11)² + (y − 5)² = 9
Centre (C₂) = (11, 5)
Radius r₂ = 3
distance (C₁C₂) =
distance (C₁C₂) = 6
∵ r₁ + r₂ = 3 + 3 = 6
∴ circles touch externally
Hence, circle have only one meeting point.

Q.44. Two tangents are drawn from a point P to the circle x² + y² − 2x − 4y + 4 = 0, such that the angle between these tangents is tan⁻¹(12/5), where tan⁻¹(12/5) ∈ (0, π). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of ΔPAB and ΔCAB is:              (JEE Main 2021)
(a) 3 : 1
(b) 9 : 4 
(c) 2 : 1 
(d) 11 : 4 

Ans. b 



In ΔCAP, 

In ΔAPM,


In ΔCAM, 

Q.45. Let the tangent to the circle x² + y² = 25 at the point R(3, 4) meet x-axis and y-axis at points P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r² is equal to:              (JEE Main 2021)
(a) 585/66
(b) 625/72
(c) 529/64
(d) 125/72

Ans. b
Given equation of circle
x² + y² = 25
∴ Tangent equation at (3, 4)
T : 3x + 4y = 25

Incentre of ΔOPQ. 

∵ Distance from origin to incenter is r.

Therefore, the correct answer is (b). 

Q.46. Choose the incorrect statement about the two circles whose equations are given below:
x² + y² − 10x − 10y + 41 = 0 and
x² + y² − 16x − 10y + 80 = 0               (JEE Main 2021)
(a) Distance between two centres is the average of radii of both the circles.
(b) Both circles pass through the centre of each other.
(c) Circles have two intersection points.
(d) Both circle's centers lie inside region of one another. 

Ans. d
S₁ ≡ x² + y² − 10x − 10y + 41 = 0
Centre C₁ ≡ (5, 5), radius r₁ = 3
S₂ ≡ x² + y² − 16x − 10y + 80 = 0
Centre C₂ ≡ (8, 5), radius r₂ = 3
Distance between centres = 3
Hence both circles pass through the centre of each other, have two intersection point and distance between two centres in average of radii of both the circles.
Hence, option (d) is the incorrect statement. 

Q.47. The line 2x − y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on x − 2y = 4. Then, the radius of the circle is:               (JEE Main 2021)
(a) 5√3
(b) 4√5
(c) 3√5
(d) 5√4

Ans. c

m₁ × m₂ = −1
a − 14 = 2 − a
2a = 16
a = 8
∴ Centre (8, 2)
Radius =
= √45
= 3√5

Q.48. Let the lengths of intercepts on x-axis and y-axis made by the circle x² + y² + ax + 2ay + c = 0, (a < 0) be 2√2 and 2√5, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to:                (JEE Main 2021)
(a) √10
(b) √6
(c) √11
(d) √7

Ans. b

a² − 4c = 8 .... (1)

a² − c = 5 .... (2)
(2) − (1)
3c = −3a ⇒ c = −1
a² = 4 ⇒ a = −2 (Given a < 0)
Equation of circle
x² + y² − 2x − 4y − 1 = 0
Equation of tangent which is perpendicular to the line x + 2y = 0 is
2x − y + λ = 0
∴ p = r

⇒ λ = ±√30
∴ Tangent 2x − y ± √30 = 0
Distance from origin

Q.49. If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x² + y² = 1 is a circle of radius r, then r is equal to:                 (JEE Main 2021)
(a) 1/4
(b) 1/2
(c) 1
(d) 1/3

Ans. b
Let P(h, k) and point on the circle is (cosθ, sinθ)

cosθ = 2h − 3 and sinθ = 2h − 2
Squaring and adding we get
(2h − 3)² + (2h − 2)² = 1
⇒ 4x² − 12x + 9 + 4y² − 8y + 4 = 1
⇒ 4x² + 4y² − 12x − 8y + 12 = 0
⇒ x² + y² − 3x − 2y + 3 = 0
 

Q.50. Let A(1, 4) and B(1, −5) be two points. Let P be a point on the circle (x − 1)² + (y − 1)² = 1 such that (PA)² + (PB)² have maximum value, then the points, P, A and B lie on:                  (JEE Main 2021)
(a) a straight line 
(b) an ellipse 
(c) a parabola 
(d) a hyperbola

Ans. a
P be a point on (x − 1)² + (y − 1)² = 1
so P(1 + cos⁡θ, 1 + sin⁡θ)
A(1, 4), B(1, −5)
(PA)² + (PB)²
= (cos⁡θ)² + (sin⁡θ − 3)² + (cos⁡θ)² + (sin⁡θ + 6)²
= 47 + 6sin⁡θ
It is maximum if sin⁡θ = 1
When sin⁡θ = 1, cos⁡θ = 0
So P(1, 2), A(1, 4), B(1, −5)
P, A, B are collinear points. 

Q.51. In the circle given below, let OA = 1 unit, OB = 13 unit and PQ ⊥ OB. Then, the area of the triangle PQB (in square units) is:                  (JEE Main 2021)
(a) 24√2
(b) 24√3
(c) 26√2
(d) 26√3

Ans. b

Let PA = AQ = λ
OA . AB = AP . AQ
⇒ 1.12 = λ . λ
⇒ λ = 2√3


= 2√4

Q.52. If the curve x² + 2y² = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is:                  (JEE Main 2021) 
(a)
(b)
(c)
(d)

Ans. d

Line : x + y = 1

Using homogenisation
x² + 2y² = 2(1)²
x² + 2y² = 2(x + y)²
x² + 2y² = 2x² + 2y² + 4xy
x² + 4xy = 0
for ax² + 2hxy + by² = 0

tan⁡θ = −4

Q.53. Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be (10/3, 7/3). If α, β are the roots of the equation ax² + bx + 1 = 0, then the value of α² + β² − αβ is:                  (JEE Main 2021) 
(a) 69/256
(b) 71/256
(c)
(d)

Ans. c
2b = a + c 

a = 4, 

b = 11/4
c = 3/2
∴ Quadratic Equation is 
 The value of
α² + β² − αβ
= α² + β² + 2αβ − 3αβ
= (α + β)² − 3αβ

Q.54. Let B be the centre of the circle x² + y² − 2x + 4y + 1 = 0. Let the tangents at two points P and Q on the circle intersect at the point A(3, 1). Then 8.is equal to _____________.                  (JEE Main 2021) 

Ans. 18


In ΔABP

AP² = AB² − BP² = 13 − 4 = 9

AP = 3

AQ = AP = 3

Let ∠ABP = θ, ∠BAP = 90 − θ

In ΔABP, tanθ = 3/2

In ΔARP, 

In ΔBRP,

cos⁡θ = BR/BP

Q.55. If the variable line 3x + 4y = α lies between the two circles (x − 1)² + (y − 1)² = 1 and (x − 9)² + (y − 1)² = 4, without intercepting a chord on either circle, then the sum of all the integral values of α is ___________.                   (JEE Main 2021)

Ans. 165

Both centers should lie on either side of the line as well as line can be tangent to circle.
(3 + 4 − α) . (27 + 4 − α) < 0
(7 − α) . (31 − α) < 0 ⇒ α ∈ (7, 31) ....... (1)
d₁ = distance of (1, 1) from line
d₂ = distance of (9, 1) from line 


(1) ∩ (2) ∩ (3) ⇒ α ∈ [12, 21]
Sum of integers = 165 

Q.56. Two circles each of radius 5 units touch each other at the point (1, 2). If the equation of their common tangent is 4x + 3y = 10, and C₁(α, β) and C₂(γ, δ), C₁ ≠ C₂ are their centres, then |(α + β) (γ + δ)| is equal to ___________.                   (JEE Main 2021) 

Ans. 40
Slope of line joining centres of circles = 4/3 = tan⁡θ

⇒ cos⁡θ = 3/5, sin⁡θ = 4/5
Now using parametric form

(x, y) = (1 + 5cosθ, 2 + 5sinθ)
(α, β) = (4, 6)
(x, y) = (γ, δ) = (1 − 5cosθ, 2 − 5sinθ)
(γ, s) = (−2, −2)
⇒ |(α + β) (γ + δ)| = | 10x − 4 | = 40

Q.57. Let the equation x² + y² + px + (1 − p)y + 5 = 0 represent circles of varying radius r ∈ (0, 5]. Then the number of elements in the set S = {q : q = p² and q is an integer} is __________.                   (JEE Main 2021) 

Ans. 61

Since, r ∈ (0, 5]
So, 0 < 2p² − 2p − 19 ≤ 100

so, number of integral values of p² is 61.

Q.58. The locus of a point, which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d² is equal to _____________.                   (JEE Main 2021) 

Ans. 16
Let point P(x, y)
A(0, 0), B(1, 0), C(0, 1), D(1, 1)
(PA)² + (PB)² + (PC)² + (PD)² = 18
x²  + y²  + x²  + (y − 1)²  + (x − 1)²  + y²  + (x − 1)²  + (y − 1)² = 18
⇒ 4(x²  + y² ) − 4y − 4x = 14


⇒ d² = 16

Q.59. The minimum distance between any two points P₁ and P₂ while considering point P₁ on one circle and point P₂ on the other circle for the given circles' equations
x² + y