Circle | JEE Advanced - Free MCQ Practice Test with solutions,

The given circle is x² + y² – 2x + 4y + 3 = 0. Centre (1, – 2). Lines through centre (1, – 2) and parallel to axes are x = 1 and y = – 2.

Let the side of square be 2k.
Then sides of square are x = 1 – k and x = 1 + k and y = – 2 – k and y = – 2 + k
∴ Co-ordinates of P, Q, R, S are (1 + k, – 2 + k), (1 – k, – 2 + k), (1 – k, – 2 – k), (1 + k, – 2  – k) respectively.
Also P (1 + k, – 2 + k) lies on circle
∴ (1 + k)² + (– 2 + k)² – 2 (1 + k) + 4 (–2 + k) + 3 = 0 ⇒ 2k² = 2  
⇒ k = 1  or – 1 If k = 1,  P (2, – 1), Q (0, – 1), R (0, – 3), S (2, – 3)
If k = – 1,  P (0, – 3), Q (2, – 3), R (2, – 1), S(0, – 1)

The circle through points of intersection of the two circles x² + y² – 6 = 0
and  x² + y² – 6x + 8 = 0
 is (x² + y² – 6) + l (x² + y² – 6x + 8) = 0
As it passes through (1, 1) (1 + 1 – 6) + l (1 + 1 – 6 + 8) = 0  ⇒ 

∴ The required circle is 2x² + 2y² – 6x + 2 = 0  
or,    x² + y² – 3x + 1 = 0

Let C (h, k) be the centre of circle touching x² = y at B (2, 4). Then equation of common tangent at B is

i.e., 4x – y = 4

Radius is perpendicular to this tangent

 = -1 ⇒ 4k = 18… (1)        

Also AC = BC ⇒ h² + (k – 1)² = (h – 2)² + (k – 4)²
⇒ 4h + 6k = 19 … (2)

Solving (1) and (2) we get the centre as  

KEY CONCEPT
Circle through pts. of intersection of two circles S₁ = 0 and S₂ = 0 is S₁ + λS₂ = 0
∴ Req. circle is, (x² + y² + 13x – 3y) + λ (x² + y² + 2x 

⇒ (1 + λ) x² + (1 +λ) y² + (13 + 2λ)

As it passes through (1, 1)

∴ 1 + λ + 1 + λ + 13 + 2λ – 3 –

⇒ – 12λ + 12 = 0   ⇒ λ = 1

∴ Req. circle is,

2x² + 2y² + 15 x - 

or4x² + 4y² + 30x –13 y – 25 = 0

Let AB be the chord with its mid pt M (h, k ).

As ∠AOB = 90°

∴ AM =

NOTE THIS STEP
By prop. of rt. Δ AM = MB = OM

∴ OM =  ⇒ h² +k² = 2

∴ locus of (h, k) is x² + y² = 2

KEY CONCEPT
Two circles x² + y² + 2g₁x + 2f₁y + c₁ = 0 and
x² + y² + 2g₂ x + 2f₂ y + c₂ = 0 are orthogonal iff
2g₁g₂ + 2f₁ f₂ = c₁ + c₂

(a) Let the required circle be, x² + y² + 2gx + 2fy + c = 0 … (1)
As it passes through (a, b), we get,
a² + b² + 2ag + 2bf + c = 0 … (2)
Also (1) is orthogonal with the circle, x² + y² = k² … (3)
For circle (1) g₁ = g, f₁ = f, c₁ = c
For circle (3) g₂ = 0, f₂ = 0, c₂ = – k²
∴ By the condition of orthogonality,
2g₁g₂ + 2f₁f₂ = c₁ + c₂ We get, c = k²
Substituting this value of c in eq. (2), we get
a² + b² + 2ga + 2f b + k² = 0
∴ Locus of centre (g – f) of the circle can be obtained by replacing g by –x and f  by – y
we get a² + b² – 2ax – 2by + k² = 0
i.e.2ax + 2by – (a² + b² + k²) = 0

We have two circles (x – 1)² + (y – 3)² = r²
Centre (1, 3), radius  = r and x² + y² – 8x + 2y + 8 = 0

Centre (4, – 1), radius =

As the two circles intersect each other in two distinct points we should have

C₁ C₂ < r₁ + r₂ and C₁ C₂ > | r₁ – r₂ |

⇒ C₁ C₂ < r + 3 and C₁ C₂ < | r₁ – r₂|

and 5 > |r – 3 |

⇒ 5 < r + 3 ⇒ | r – 3 | < 5        ⇒ r > 2 … (i)

⇒ – 5 < r – 3 < 5              ⇒ – 2 < r < 8 … (ii)

Combining (i) and (ii), we get 2 < r < 8

As 2x – 3y – 5 = 0 and 3x – 4y – 7 = 0 are diameters of circles.

∴ Centre of circle is solution of these two eq. ' s, i.e.

⇒ x = 1, y = – 1

∴ C (1, –1)
Also area of circle, πr² = 154

∴ Equation of required circle is

(x – 1)² + (y + 1)² = 72  

⇒ x² + y² – 2x + 2y = 47

Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0.
As this circle passes through (0, 0) and (1, 0)
we get c = 0, 1 + 2g = 0

⇒ g =

According to the question this circle touches the given circle x² + y² = 9

∴ 2 × radius of required circle = radius of given circle

∴ The centre is 

The given circle is x² + y² – 6x + 14 = 0, centre (3, 3), radius = 2
Let (h, k) be the centre of touching circle.

Then radius of touching circle = h [as it touches y-axis also]

∴ Distance between centres of two circles =  sum of the radii of two circles

⇒ (h – 3)² + (k – 3)² = (2 + h)²

⇒ k² – 10h – 6k + 14 = 0

∴ locus of (h, k) is  y² – 10x – 6y + 14 = 0

Centres and radii of two circles are C₁ (5, 0); 3 = r₁ and C₂ (0, 0); r = r₂

As circles intersect each other in two distinct points,

|r₁ – r₂| < C₁C₂ < r₁ + r₂

⇒ | r – 3 | < 5 < r + 3 ⇒ 2 < r < 8

Centre of the circle x² + y² + 4x – 6y + 9 sin²α + 13 cos²α = 0 is C (– 2, 3) and its radius is

Let P (h, k) be any point on the locus. The ∠ APC =α

Also ∠ PAC =

That is, triangle APC is a right triangle.

Thus, 

⇒ (h + 2)² + (k – 3)² = 4 or    h² + k² + 4h – 6k + 9 = 0

Thus required equation of the locus is x² + y² + 4x – 6y + 9 = 0

The given equation of the circle is x² + y² – px – qy  = 0, pq ≠ 0 Let the chord drawn from (p, q) is bisected by x-axis at point (x1, 0).
Then equation of chord is

(using T = S₁)

As it passes through (p, q), therefore,

As through (p,q) two distinct chords can be drawn.

∴ Roots of above equation be real and distinct,  i.e., D > 0.

⇒ 9p² – 4 × 2 (p² + q²) > 0 ⇒ p² > 8q²

O is the point at centre and P is the point at circumference. Therefore, angle QOR is double the angle QPR.

So, it sufficient to find the angle QOR. Now slope of OQ = 4/3
Slope of OR = – 3/4
Again m₁ m₂ = – 1
Therefore, ∠QOR = 90° which implies that ∠QPR = 45°.

2g₁g₂ + 2f₁f₂ = c₁+ c₂ (formula for orthogonal intersection of two cricles)

⇒ 2 (1) (0) + 2 (k) (k) = 6 + k

⇒ 2k² – k – 6 = 0  

⇒ k = – 3/2, 2

x² + y² = r² is a circle with centre at (0, 0) and radius r units.

Any arbitrary pt P on it is (r cosθ , r sinθ ) Choosing A and B as (– r, 0) and (0, – r).

[So that ∠AOB = 90°]

For locus of centriod of ΔABP

⇒ r cosθ – r = 3x
r sinθ – r = 3y

⇒ r cosθ = 3x + r
r sinθ = 3y + r

Squaring and adding (3x + r)² + (3y + r)² = r² which is a circle.

Let ∠RPS = θ

∠XPQ = 90° – θ

∴∠PQX = θ (∵ ∠PXQ = 90°)

∴ΔPRS ~ ΔQPR (AA similarity)

Line 5x – 2y + 6 = 0 is intersected by tangent at P to circle x² + y² + 6x + 6y – 2 = 0 on y-axis at Q (0, 3).
In other words tangent passes through (0, 3)
∴ PQ = length of tangent to circle from (0, 3)

x² – 8x + 12 = 0 ⇒ (x – 6) (x – 2) = 0
y² – 14y + 45 = 0 ⇒ (y – 5) (y – 9) = 0
Thus sides of square are x = 2, x = 6, y = 5, y = 9
Then centre of circle inscribed in square will be

The given circle is x² + y² – 2x – 6y + 6 = 0 with centre C(1, 3) and radius

 Let AB be one of its diameter which is the chord of other circle with centre at C₁ (2, 1).
Then in ΔC₁CB,

r² = [(2 – 1)² + (1 – 3)²] + (2)²

⇒ r² = 1 + 4 + 4  ⇒ r² = 9  ⇒ r = 3.

Let the centre of circle C be (h, k). Then as this circle touches axis of x, its radius = | k |

Also it touches the given circle x² + (y – 1)² = 1, centre (0, 1) radius 1, externally

Therefore, the distance between centres = sum of radii

⇒ h² + k² – 2k + 1 = (1 + | k |)²
⇒ h² +  k² – 2k + 1 = 1 + 2 | k | + k²
⇒ h² = 2k + 2 | k |
∴ Locus of (h, k) is, x² = 2y + 2 | y |
Now if y > 0, it becomes x² = 4y and if y < 0, it becomes x = 0
∴ Combining the two, the required locus is {(x, y) : x² = 4y} ∪ {(0, y) : y < 0}

Tangents PA and PB are drawn from the point P (1, 3) to circle x² + y² - 6 x - 4y - 11=0 with centre C (3, 2)

Clearly the circumcircle of ΔPAB will pass through C and as ∠A = 90°, PC must be a diameter of the circle.

∴ Equation of required circle is (x -1)(x - 3) + (y-8)(y - 2)= 0

⇒ x² + y² - 4x - 10y + 19= 0

Let centre of the circle be (h, 2) then radius = |h|

∴ Equation of circle becomes (x – h)² + (y – 2)² = h²

As it passes through (–1, 0)

⇒ (–1 – h)² + 4 = h² ⇒ h =  

Distance of centre from (–4, 0) is  

∴ It lies on the circle.

Any point P on line  4x – 5y = 20  is 

Equation of chord of contact AB to the circle x² + y² = 9

drawn from point 

....(1)

Also the equation of chord AB whose mid point is (h, k) is

hx + ky = h²+k² ....(2)

∵  Equations (1) and (2) represent the same line, therefore

⇒ 5kα= 4hα- 20h and 9h = α( h²+k²)

and 

∴ Locus of (h, k) is

(20 x² + y²) - 36x + 45y= 0