JEE Advanced (Fill in the Blanks): Circle | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. If A and B are points in the  plane such that PA/PB = k (constant) for all P on a given circle, then the value  of k cannot be equal to ..................... (1982 - 2 Marks) 

Ans. 1

Sol. As P lies on a circle and A and B two points in the plane

such that  

Then k can be any real number except 1 as otherwise P will lie on perpendicular bisector of AB which is a line.

Q.2. The points of intersection of the line 4x – 3y – 10 = 0 and the circle x² + y² – 2x + 4y – 20 = 0 are ....................  and ..................... (1983 - 2 Marks) 

Ans.  (4, 2), (–2, –6)

Sol. For point of intersection of line 4x – 3y – 10 = 0 … (1)
and circle x² + y² – 2x + 4y – 20 = 0 … (2)
Solving (1) and (2), we get

⇒ y² + 4y – 12 = 0  ⇒ y = 2, – 6
⇒ x = 4,– 2
∴ Points are (4, 2) and  (– 2, – 6)

Q.3. The lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0  are tangents to the same circle. The radius of this circle is ..................... (1984 - 2 Marks)

Ans. 3/4

Sol. Let 3x – 4y + 4 = 0 be the tangent at point A and 6x – 8y – 7 = 0 be the tangent of point B of the circle.
As the two tangents parallel to each other
∴ AB should be the diameter of circle.
∴ AB = distance between parallel lines
3x – 4y + 4 = 0 and
6x – 8y – 7 = 0 = distance between
6x – 8y + 8 = 0 and     6x – 8y – 7 = 0

∴ radius of circle =

Q.4. Let x² + y² – 4x – 2y – 11 = 0 be a circle. A pair of tangents from the point (4, 5) with a pair of radii form a quadrilateral of area ..................... (1985 - 2 Marks) 

Ans. 8 sq. units 

Sol. KEY CONCEPT :
Length of tangent from a point (x₁, y₁) to a circle x² + y² + 2gx + 2fy + c = 0 is given by

The equation of circle is, x² + y² – 4x – 2y – 11 = 0
It's centre is (2, 1), radius =

length of tangent  from the pt. (4, 5) is

∴ Area of quad. ABCD

= 2 (Area of ΔABC)

 = 8 sq. units.

Q.5. From the origin chords are drawn to the circle (x – 1)² + y² = 1. The equation of the locus of the mid-points of these chords is ..................... (1985 - 2 Marks) 

Ans. x² + y² – x = 0

Sol. The equation of given circle is (x – 1)² + y² = 1
or x² + y² – 2x = 0 … (1)

KEY CONCEPT : We know that equation of chord of curve S = 0, whose mid point is (x₁, y₁) is given by T = S₁ where T is tangent to curve S = 0 at (x₁, y₁).
∴ If (x₁, y₁) is  the mid point of chord of given circle (1), then equation of chord is

At it passes through origin, we get

Q.6. The equation of the line passing through the points of intersection of the circles 3x² + 3y² – 2x + 12y – 9 = 0 and x² + y² + 6x + 2y – 15 = 0 is ..................... (1986 - 2 Marks) 

Ans. 10x – 3y – 18 = 0

Sol. The equation of two circles are

… (1)

and x² + y² + 6x + 2y – 15 = 0 … (2)

Now we know eq. of common chord of two circles

S₁ = 0 and S₂ = 0 is S₁ – S₂ = 0

10x – 3y – 18 = 0

Q.7. From the point A(0, 3) on the circle x² + 4x + (y – 3)² = 0, a chord AB is drawn and extended to a point M such that AM = 2AB. The equation of the locus of M is .................. (1986 - 2 Marks) 

Ans. 

Sol. The equation of circle is, x² + y² + 4x – 6y + 9 = 0 … (1)

AM = 2AB
⇒ AB = BM
Let the co-ordinates of M be (h, k) Then B is mid pt of AM

As B lies on circle (1),

⇒ h² + k² + 8h – 6k + 9 = 0
∴ locus of (h, k) is, x² + y² + 8x – 6y + 9 = 0

Q.8. The area of the triangle formed by the tangents from the point (4, 3) to the the circle x² + y² = 9 and the line joining their points of contact is ..................... (1987 - 2 Marks) 

Ans. 192/25

Sol. From P (4, 3) two tangents PT and PT' are drawn to the circle x² + y² = 9 with O (0, 0) as centre and r = 3.
To find the area of ΔPTT'.

Let R be the point of intersection of OP and TT'.
Then we can prove by simple geometry that OP is perpendicular bisector of TT'.
Equation of chord of contact TT' is 4x + 3y = 9 Now, OR =  length of the perpendicular from O to TT' is

OT = radius of circle = 3

Again OP =  

∴ PR = OP - OR =  

Area of the triangle

PTT' = PR x TR

Q.9. If the circle C₁ : x² + y² = 16 intersects anoth er circle C₂ of radius 5 in such a manner that common chord is of maximum length and has a slope equal to 3/4, then the coordinates of the centre of C₂ are .....................           (1988 - 2 Marks) 

Ans. 

Sol. We have C₁ : x² + y² = 16, Centre O₁ (0, 0) radius  = 4. C₂ is another circle with radius 5, let its centre O₂ be (h, k).

Now the common chord of circles C₁ and C₂ is of maximum length when chord is diameter of smaller circle C₁, and then it passes through centre O₁ of circle C₁. Given that slope of this chord is 3/4.

∴ Equation of AB is,

… (1)

In right ΔAO₁O_2,

Also  distance from (h, k) to (1)

⇒ 3h – 4k ± 15 = 0 … (2)

… (3)

Solving, 3h – 4k + 15 = 0 and 4h + 3k = 0
We get h = – 9/5, k = 12/5
Again solving 3h – 4k – 15 = 0 and 4h + 3k = 0
We get h = 9/5, k = –12/5

Thus the required centre is  

Q.10. The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle x² + y² = 4 at  is, ..................... (1989 - 2 Marks) 

Ans. 

Sol. Tangent at P (1, ) to the circle x² + y² = 4  is x . 1 + y . = 4

It meets x-axis at A (4, 0),

∴ OA = 4 Also OP = radius of circle = 2  

∴ Area of ΔOPA =

 sq. units

Q.11. If a circle passes through the points of intersection of the coordinate axes with the lines λx – y + 1 = 0 and x – 2y + 3 = 0, then the value of λ =  ................... (1991 -  2 Marks) 

Ans.  2

Sol. The given lines are lx – y + 1 = 0 and x – 2y + 3 = 0 which meet x-axis at   and B (– 3, 0) and
y-axis at C (0, 1) and D    respectively..
Then we must have, OA × OB = OC × OD

Q.12. The equation of the locus of the mid-points of the circle 4x² + 4y² – 12x + 4y + 1 = 0 that subtend an angle of 2π/ 3 at its centre is ..................... (1993 -  2 Marks)

Ans. 16x² + 16y² – 48x + 16y + 31 = 0

Sol. The given circle is, 4x² + 4y² – 12x + 4y + 1 = 0

or with centre  

and

Let M (h, k) be the mid pt. of the chord AB of the given circle, then CM ⊥ AB. ∠ ACB = 120°.
In ΔACM,

and ∠A = 30°

⇒ 16h² + 16k² – 48h + 16k + 31 = 0
∴ locus of (h, k) is 16x² + 16y² – 48x + 16y + 31 = 0

Q.13. The intercept on the line y = x by the circle x² + y² – 2x = 0 is AB. Equation of the circle with AB as a diameter is ..................... (1996 - 1 Mark) 

Ans. x² + y² – x – y = 0

Sol. Equation of any circle passin g through the point of intersection of x² + y² – 2x = 0 and  y
= x is x² + y² – 2x + l (y – x) = 0
or x² + y² – (2 + l)x + ly = 0

Its centre is

For AB to be the diameter of the required circle, the centre must lie on AB. That is,

Thus, equation of required circle is x² + y² – 2x – y + x = 0 or x² + y² – x – y = 0

Q.14. For each natural number k, let C_k denote the circle with radius k centimetres and centre at the origin. On the circle C_k, a-particle moves k centimetres in the counter-clockwise direction. After completing its motion on C_k, the particle moves to C_k+1 in the radial direction. The motion of the particle continues in this manner. The particle starts at (1, 0).
 If the particle crosses the positive direction of the x-axis for the first time on the circle C_n then n = ..................... (1997 - 2 Marks) 

Ans. 7

 Sol. 

The radius of circle C₁ is 1 cm, C₂ is 2 cm and soon.
It starts from A₁ (1, 0) on C₁, moves a distance of 1 cm on C₁ to come to B₁. The angle subtended by A₁B₁ at the centre

will be radians, i.e. 1 radian.

From B1 it moves along radius, OB₁ and comes to A₂ on circle C₂ of radius 2. From A₂ it moves on C₂ a distance 2 cm and comes to B₂. The angle subtended by A₂B₂ is again as before 1 radian. The total angle subtended at the centre is 2 radians. The process continues. In order to cross the x-axis again it must describe 2p radians i.e  radians.
Hence it must be moving on circle C₇

∴ n = 7

Q.15. The chords of contact of the pair of tangents drawn from each point on the line 2x+y =4 to circle x2+y2 = 1 pass through the point ..................... (1997 - 2 Marks)

Ans. 

Sol. Let (h, k) be any point on the given line
∴ 2h + k = 4 and chord of  contact is hx + ky  = 1 or hx + (4 – 2h) y = 1
or (4y – 1) + h (x – 2y) = 0
P + l Q = 0.It passes through the intersection of P = 0 and