JEE Advanced Previous Year Questions (2018 - 2023): Circle | Mathematics (Maths) for JEE Main & Advanced PDF Download

2023

Q1: Let A₁, A₂, A₃, …, A₈ be the vertices of a regular octagon that lie on a circle of radius 2 . Let P be a point on the circle and let PA_i denote the distance between the points P and A_i for i = 1 ,2,…,8. If P varies over the circle, then the maximum value of the product PA₁ × PA₂ × ⋯⋯ × PA₈, is :    [JEE Advanced 2023 Paper 2]
Ans: 512
Q2: Let C₁ be the circle of radius 1 with center at the origin. Let C₂ be the circle of radius r with center at the point A = (4, 1), where 1< r <3. Two distinct common tangents PQ and ST of C₁ and C₂ are drawn. The tangent PQ touches C₁ at P and C₂ at Q. The tangent ST touches C₁ at S and C₂ at T. Mid points of the line segments PQ and ST are joined to form a line which meets the x-axis at a point B. If AB = √5, then the value of r² is :                   [JEE Advanced 2023 Paper 2]
Ans: 2

Q3: Consider an obtuse angled triangle ABC in which the difference between the largest and the smallest angle is π/2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Let  be the area of the triangle . Then the value of  is  Let a be the area of the triangle ABC. Then the value of (64a)² is  :   [JEE Advanced 2023 Paper 2]
Ans: 1008

Q4: Consider an obtuse angled triangle ABC in which the difference between the largest and the smallest angle is π/2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Then the inradius of the triangle ABC is  :    [JEE Advanced 2023 Paper 2]
Ans: 0.25

2022

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,

= 0.837≅ 0.84 

2021

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).

Q2: 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 Paper 2]
Ans: 2.00
Given, x ≥ 0, y² ≤ 4 − x
Let equation of circle be
(x − h)² + y² = h² .... (i) 2

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 

Q3: Let , 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
(a) k + 2l = 22
(b) 2k + l = 26
(c) 2k + 3l = 34
(d) 3k + 2l = 40                [JEE Advanced 2021 Paper 2]
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 

Q4: Let , 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 . The number of all those circles D_n that are inside M is 
(a) 198
(b) 199
(c) 200
(d) 201                [JEE Advanced 2021 Paper 2]
Ans: (b)

Now,

n ≤ 199
So, number of circles = 199 

Q5: 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 Paper 1]
(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 
x² + y² + 2 × 1/2x + 2 × 3/2y = 0
i.e. x² + y² + x + 3y = 0 

2020

2019

From the figure,

and

∵ AB = AM + MB = 2AM [∵ AM = MB] 

From Eqs. (ii) and (iii), we get

From the Eq. (i), we get
AC = 2/sinθ = 2 x 5 = 10

Q2: A line y = mx + 1 intersects the circle (x − 3)² + (y + 2)² = 25 at the points P and Q. If the midpoint of the line segment PQ has x-coordinate −3/5, then which one of the following options is correct?
(a) 6 ≤ m < 8
(b) −3 ≤ m < −1
(c) 4 ≤ m < 6
(d) 2 ≤ m < 4                    [JEE Advanced 2019 Paper 1]
Ans: (d)
It is given that points P and Q are intersecting points of circle (x − 3)² + (y + 2)² = 25 .....(i)
Line y = mx + 1 .....(ii)
And, the mid-point of PQ is A having x-coordinate −3 / 5
so y-coordinate is 1 − 3 / 5 m.
So, 
From the figure,
∵ AC ⊥ PQ 

⇒ (slope of AC) × (slope of PQ) = −1 

⇒ 3m² - 15m + 18 = 0
⇒ m² - 5m + 6 = 0
⇒ m = 2 or 3