Solved Objective Question on Circle Set 1

Q1.

Tangents are drawn to the circle x² + y² = 50 from a point ‘P’ lying on the x-axis. These tangents meet the y-axis at points ‘P₁’ and ‘P₂’ . Possible coordinates of ‘P’ so that area of triangle PP₁P₂ is minimum, is / are.

(A) (10, 0)

(B) ( 10, 0)

(D) ( -10, 0)

Solution:

OP = 5secq,

OP₁ = 5cosecq

D PP₁P₂ =

(D PP₁P₂)_min = 100

&⇒ q = p/4
&⇒ OP =10

&⇒ P=(10, 0), (-10, 0) .

Hence (A), (C) are correct

Q2.

Two circles with radii ‘r₁’ and ‘r₂’, r₁ > r₂ ≥ 2 , touch each other externally. If ‘q’ be the angle between the direct common tangents, then .

(A)

(B)

(D) none of these.

Solution:

sina =

&⇒ q = 2 sin^-1

Hence (B) is correct.

Q3.

If the curves ax²+4 xy+2y²+x+y+5 =0 and ax²+6xy+5y²+ 2 x+ 3y + 8 = 0 intersect at four concyclic points then the value of a is

(A) 4

(B) -4

(D) –6

Solution:

Any second degree curve passing through the intersection of the given curves is

ax² + 4xy + 2y² + x + y + 5 + l ( ax² + 6xy + 5y² +2 x + 3y + 8 ) = 0

If it is a circle, then coefficient of x² = coefficient of y² and coefficient of xy = 0

a(1+ l) = 2 + 5l and 4 + 6 l = 0

&⇒ a = and l =
&⇒ a = = – 4 .

Hence (B) is correct answer.

Q4.

The chords of contact of the pair of tangents drawn from each point on the line 2x + y = 4 to the circle x² + y² =1 pass through a fixed point

(A) (2 , 4)

(B)

(C)

(D) ( -2, -4)

Solution:

The chord of contact of tangents from (a, b) is

ax + b y = 1 . . . . . (1).

Also, (a, b )lies on 2x +y = 4 , so 2a + b = 4

&⇒

Hence, (1) passes through .

Hence (C) is correct answer.

Q5.

Equation of chord AB of circle x² + y² = 2 passing through P(2 , 2) such that PB/PA = 3, is given by

(A) x = 3 y

(B) x = y

(D) none of these

Solution:

Any line passing through (2, 2) will be of the form = r

When this line cuts the circle x²+y²=2 , (rcosq+2)² +(r sinq+2)² =2

&⇒ r² + 4(sinq+ cosq)r +6 = 0

, now if r₁ = a, r₂ = 3a,

then 4a = - 4(sinq + cosq), 3a² = 6
&⇒ sin2q = 1
&⇒ q = p/4 .

So required chord will be y – 2 = 1 ( x –2)
&⇒ y = x.

Alternative solution

PA.PB = PT² = 2² + 2² – 2 = 6 . . . . (1).

. . . . (2)

From (1) and (2), we have PA = , PB =3

&⇒ AB = 2 . Now diameter of the circle is 2 (as radius is)

Hence line passes through the centre
&⇒ y = x .

Hence (B) is the correct answer. .

Q6.

Equation of a circle S(x , y) = 0, (S(2, 3) = 16) which touches the line 3x+ 4y –7 = 0 at (1, 1) is given by

(A) x² +y² +x +2y –5 =0

(B) x² +y² +2x +2y –6 =0

(D) none of these

Solution:

Any circle which touches 3x +4y – 7 =0 at (1, 1) will be of the form

S(x, y) º (x –1)² + (y-1)² +l(3x +4y-7) = 0

Since S(2, 3) = 16
&⇒ l =1, so required circle will be

x² +y² +x +2y –5 =0.

Hence (A) is the correct answer. .

Q7.

If (a, 0) is a point on a diameter of the circle x²+y² =4, then
x² – 4x – a² = 0 has

(A) exactly one real root in ( –1, 0]

(B) exactly one real root in [ 2, 5]

(D) distinct roots less than 5

Solution:

Since (a, 0) is a point on the diameter of the circle

x² +y² = 4,

so maximum value of a² is 4

Let f(x) = x² – 4x – a²
clearly f(-1) = 5 – a² > 0,

f(2) = -(a²+ 4) < 0

f(0)= -a²<0 and f( 5)= 5- a² > 0

so graph of f(x) will be as shown

Hence (A), (B), (C), (D) are the correct answers.

Q9.

If a circle S(x , y) = 0 touches at the point (2, 3) of the line x +y = 5 and S(1, 2) = 0, then radius of such circle

(A) 2 units

(B) 4 units

(D) units

Solution:

Desired equation of the circle is (x –2)² + (y –3)² + l( x +y –5) = 0

1 +1 + l (1+ 2 – 5 ) = 0
&⇒ l =1

x² – 4x + 4 + y² – 6y + 9 + x + y –5 = 0
&⇒ x² + y² – 3x – 5y + 8 = 0

Hence (D) is the correct answer.

Q9.

If P(2, 8) is an interior point of a circle x² + y² –2x + 4y – p = 0 which neither touches nor intersects the axes, then set for p is

(A) p < -1

(B) p < -4

(D) f

Solution:

For internal point p(2, 8), 4 + 64 – 4 + 32 – p < 0
&⇒ p > 96

and x intercept = 2 therefore 1 + p < 0

&⇒ p < -1 and y intercept = 2
&⇒ p < -4

Hence (D) is the correct answer.

Q10.

If two circles (x – 1)²+ (y – 3)² = r² and x²+ y²– 8x + 2y + 8 = 0 intersect in two distinct points then

(A) 2 < r < 8

(B) r < 2

(D) r > 2

Solution:

Let d be the distance between the centres of two circles of radii r₁
and r₂ . .

These circle intersect at two distinct points if ½r₁-r₂ ½ < d < r₁+r₂

Here, the radii of the two circles are r and 3 and distance between the

centres is 5.

Thus, ½r-3½ < 5 < r+3
&⇒ -2 < r < 8 and r > 2
&⇒ 2 < r < 8.

Hence (A) is the correct answer. .

Q11.

The common chord of x²+y²-4x-4y = 0 and x²+y² = 16 subtends at the origin an angle equal to

(A) p/6

(B) p/4

(D) p/2

Solution:

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.

Q12.

The number of common tangents that can be drawn to the circle
x²+y²–4x-6y-3 = 0 and x²+y²+2x+2y+1=0 is

(A) 1

(B) 2

(D) 4

Solution:

The two circles are

x²+ y²– 4x – 6y – 3 = 0 and x²+y²+2x+2y+1 = 0

Centre: C₁ º (2, 3), C₂ º (–1, –1), radii: r₁ = 4, r₂ = 1

We have, C₁ C₂ = 5 = r₁ + r₂, therefore there are 3 common tangents to the given circles. Hence (C) is the correct answer.

Q13.

The tangents drawn from the origin to the circle x²+y²-2rx-2hy+h² = 0 are perpendicular if

(A) h = r

(B) h = -r

(D) r²= h²

Solution:

The combined equation of the tangents drawn from (0,0) to

x²+ y²– 2rx – 2 hy + h² = 0 is

(x²+ y² – 2 rx – 2 hy + h²)h² = ( – rx – hy + h²)²

This equation represents a pair of perpendicular straight lines

If Coeff. of x² + coeff. of y² = 0 i.e. 2h² – r² – h² = 0 .

&⇒ r² = h² or r = ± h. Hence (A), (B), and (D) are correct answers.

Q14.

The equation(s) of the tangent at the point (0, 0) to the circle, making intercepts of length 2a and 2b units on the coordinate axes, is (are)

(A) ax + by = 0

(B) ax – by = 0

(D) None of these

Solution:

Equation of circle passing through origin and cutting off intercepts 2a and 2b units on the coordinate axes is x² + y² ± 2ax ± 2by = 0

and equation of tangent at (0, 0) is ax ± by = 0

Hence (A), (B) are correct answers.

Q15.

The slope of the tangent at the point (h,h) of the circle x²+ y²= a² is

(A) 0

(B) 1

(D) Depend on h

Solution:

The equation of the tangent at (h, h) to x² + y² = a² is hx + hy = a².

Therefore slope of the tangent = -h/h = -1

Hence (C) is the correct answer.