Test: Circles - 4 - JEE MCQ

(i.e. angle subtended at the centre of a circle is double the angle subtended in the alternate segment).
Hence (c) is the correct answer.

Since the circles touch each other, equation of the common tangent is S₁ – S₂ = 0 ⇒ x = 0.
Putting x = 0 in the equations of the circles, we get y²+ b = 0.
This equation should have equal roots ⇒ b = 0.
Hence (c) is the correct answer.

Hence (B) is the correct answer.

The equation of the common chord of the circles x²+y²-4x-4y = 0 and x²+y² = 16 is x + y = 4 which meets the circle x2+y2 = 16 at points A(4,0) and B(0,4). Obviously OA ⊥ OB. Hence the common chord AB makes a right angle at the centre of the circle x²+y² = 16.
Hence (d) is the correct answer.

(A) The  radical axis of the circle x² + y² = 1 and the circle given in is x = -2, which is | | to y-axis 
(B) is y = 2, which is | | to x-axis
(C) is x = 0, which is y-axis 
The radical axis with circle in (A) | | to y-axis and the given circle intersect circle given in A orthogonally. 

Radius of the given circle = |g|. For perpendicular tangents origin should lie on the director circle, i.e. distance of origin from the centre of the given circle  

Centre is (1, 3) and radius = 2
If r = radius of second circle then r² = 2² + (3 - 1)² + (2 - 1)²
⇒ r = 3.
Hence (C) is the correct answer.

Hence (C) is the correct answer.

Since (k + 1, k) lies inside the region bounded by x 

⇒ (k + 1)² + k² - 25 < 0
and k + 1 > 0
⇒ 2k² + 2k - 24 < 0 and k > -1
⇒ -4 < k < 3 and k > -1
⇒ -1 < k < 3
Hence (A) is the correct answer.

Since AB is the diameter of the circle. 
From figure, 
Also equation of the circle is 

equation of tangent to this circle at (0, 0) is ax + by = 0 ……(1)
∴ m = length of perpendicular from


and n = length of perpendicular from 


Hence (D) is the correct answer.

Let the centre be C (h, k). Since the circle passes through the origin,

Let x = c meets the circle in A and B.

From figure, CB² = CD² + BD²
⇒ OC² = CD² + b²  (∴CB=OC and BD=b)
⇒ h² + k² = (h - c)² + b²
⇒ k² + 2ch - c² - b² = 0
⇒ locus is y² + 2cx = b² + c².
Hence (C) is the correct answer.

Any second degree curve passing through the intersection of the given curves is ax² + 4xy + 2y² + x + y + 5 + λ (ax² + 6xy + 5y² +2x + 3y + 8) = 0
If it is a circle, then coefficient of x² = coefficient of y² and  coefficient of xy = 0 a(1+ λ) = 2 + 5λ and   4 + 6λ = 0


Hence (B) is the correct answer.

Any line passing through (2, 2)  will be  of  the  form 
When this line  cuts  the  circle  x²+y²=2 , (rcosθ+2)² +(rsinθ+2)² = 2 
⇒ r² + 4(sinθ + cosθ)r+6 = 0
now if r₁ = α, r₂ = 3α then 4α
=  - 4(sinθ + cosθ), 3α² = 6 ⇒ sin2θ = 1⇒ θ = π/4 .
So required chord will be y – 2  = 1 (x –2) ⇒ y = x.
Alternative solution 
PA.PN = PT² = 2² + 2² - 2 = 6
PB/PA = 3
From (1) and (2), we have PA = √2, PB = 3 √​2
⇒ AB = 2 √​2
Now diameter of the circle is 2 √​2(as radius is √​2) 
Hence line passes through the centre ⇒ y = x .
Hence (B) is the correct answer. 

For internal point p(2, 8),  4 + 64 – 4 + 32 – p < 0 ⇒ p > 96 
and x intercept = therefore 1 + p < 0
⇒ p < -1 and  y intercept = ⇒ p < -4
Hence (D) is the correct answer.

⇒ |r₂ - r₂| is minimum when x is minimum.
|r₂ - r₂| = 1

The point (2 cosθ, 2 sinθ) lies on the circle x² + y2 = 4. From the figure, it  is  obvious that 



Hence no solution

The radical axis of given circles is 

This line is tangent to the circle x² + y² – 4x – 4y + 4 = 0
⇒ either g = 1 or f = 1/4

Clearly (-g, -f) will be the centre of required circle, 
Let the circle be (x + g)² + (y + f)² = d
⇒ x² + y² + 2gx + 2fy + g² + f² –d = 0
It has to cut the given circle orthogonally,
⇒ 2g² + 2f² = c + g² + f² –d
⇒ d = c –g² –f²

The line 2y = gx + β should pass through (–g, –g) so,
–2g = –g² + α ⇒ α = g² – 2g = (g – 1)² – 1 > –1.

Let (h, k) be point of intersection of the tangents Then equation of chord of contact is xh + yk = 10
Compare this with x +y = 2 

Since AB is fixed and AC is perpendicular to BC. So, locus of C is a circle whose diameter is AB. So, family of circles passing through AB is 
x² + y² - 6x - 8y + 21+λ(x + y -5) = 0
⇒x² + y² - (6 - λ)x - (8 - λ)y + 21- 5λ = 0

Since AB is diameter so centre must lie on AB

∴ Locus is x² + y² - 4x - 6y + 11 = 0.

Given equation of lines 





∠APO = 75°
Length of chord of contact AB

= 6(sin45°cos30° + sin30°cos45°)

If (x_i, y_i) is the point of intersection of given curves 



⇒ y₄ = 3x₄ ⇒ locus of D is y = 3x.

 now if centre of circle is(5, 7) ≡ C say then PACB is a cyclic quadrilateral, so circumcentre of triangle PAB is midpoint of PC i.e. 



⇒ 4x² - 4xy - 6xy + 10y - 10 = 0

Shortest distance between 2 curves occurs along the common normal. Let required point be (t, 2 -t). Normal at this will be y –(2 –t) = (x –t), it should passes through centre of circle so –1 –2 + t = 5 –t ⇒ t = 4, so point can be (4, -2).