Test: 35 Year JEE Previous Year Questions: Circle

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 45^o.

⇒ 1 – m² = ± 2m ⇒ m² ± 2m – 1 = 0

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

we should have

OA ≤ OP ≤ OB ⇒ (5-3)

⇒ 4 ≤ x² +y² ≤ 64

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

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

r₁ - r₂< C₁C₂ for intersection

and ,
From (1) and (2),  2 < r < 8.

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

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 + b² + 4

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 +g² =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

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

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

s₁ = x² + y²+ 2ax + cy +a =0  
s₂ = x² + y² - 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.

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 = r₁ + r₂

∴ Locus is which is a parabola.

Let the centre be (α, β)

∵ It cuts the circle x² + y² = 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 .

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 = 45^o

∴ Angle between the diameters represented by combined equation

ax² + 2 ( a + b ) xy + by² = 0 is 45^o

∴ Using 

we get 

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

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

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

∴ Locus of (h, k) is 

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 ⇒

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)

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

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

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

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.

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 

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.

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 = 

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 

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.

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.

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