Mathematics
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Must do Problems of circle

Q 2875480366

Consider the two circles


`S_1 ≡ x^2 +y^2 -4=0` and `S_2 ≡ x^2 +y^2 - 6x-8y-24=0` .
The two circles `S_1` and `S_2`

(A)

touch each other externally

(B)

touch each other internally

(C)

cuts each other at two points

(D)

None o' the above

Solution:

Centre of circle `S_1` is `C ( 0, 0)`.

Centre of circle `S_2` is` C (3, 4)`.

`r_1 = 2` and `r_2 = sqrt (3^2 + 4^2 +24) = 7`

Now `C_1 C_2 = sqrt (3^2 + 4^2) = 5`

`C_1 C_2 <` Sum of the radii

Also `C_1 C_2 =` Difference of the radii

Thus, the two circles touch each other
internally.
Correct Answer is `=>` (B) touch each other internally
Q 2875480366

Consider the two circles


`S_1 ≡ x^2 +y^2 -4=0` and `S_2 ≡ x^2 +y^2 - 6x-8y-24=0` .
The nmr.ber of common tangents to the given
circles is

(A)

0

(B)

1

(C)

3

(D)

4

Solution:

Since, the two circles touch each
other internally. Therefore, there is only
one common tangent.
Correct Answer is `=>` (B) 1
Q 2875480366

Consider the two circles


`S_1 ≡ x^2 +y^2 -4=0` and `S_2 ≡ x^2 +y^2 - 6x-8y-24=0` .
The number of tangents from centre of circle `S_2`
to the cirde `S_1` is

(A)

0

(B)

1

(C)

2

(D)

None o' the these

Solution:

We have,

`S_1 = (3)^2 + ( 4)^2 - 4 = 21 > 0`

`:. `Point (3, 4) lies outside the circle `S_1`

`:.` Two tangents can be drawn from (3, 4).
Correct Answer is `=>` (C) 2
Q 2875480366

Consider the two circles


`S_1 ≡ x^2 +y^2 -4=0` and `S_2 ≡ x^2 +y^2 - 6x-8y-24=0` .
Consider the following statements
I. Equati:m of common chord is `3x + 4y + 10 = 0`.
II. Equati:m of common tangent is `4y + 3x = 5`.

Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

None o' the above

Solution:

I. Equation of common chord is
given by `S_1 - S_2= 0`

i.e. `- 4 + 6x + 8 y + 24 = 0`

or `3x + 4y + 10 = 0`

So, Statement I is true.

II. Since, two circles touch each other
internally, equation of common

tangent is same as equation of
common chord .
So, Statement II is false.
Correct Answer is `=>` (A) Only I
Q 2855167964

Consider the two circles

`S_1 = x^2 + y^2 - 6x + 4y + 11 = 0 `and

`S_2 = x^2 + y^2 - 4x + 6y + 9 = 0`.
The equation of common chord is

(A)

`y - x = 1`

(B)

`x- y = 1`

(C)

`x+ y -1 = 0`

(D)

`x+ y = 1`

Solution:

Equation of common chord is given by

`S_1 - S_2 = 0`

`=> - 6x+ 4y+ 11-(-4x + 6y + 9) = 0`

`=> - 2x - 2 y + 2 = 0 => x + y - 1 = 0`
Correct Answer is `=>` (C) `x+ y -1 = 0`
Q 2855167964

Consider the two circles

`S_1 = x^2 + y^2 - 6x + 4y + 11 = 0 `and

`S_2 = x^2 + y^2 - 4x + 6y + 9 = 0`.
The angle of intersection of the two circle is

(A)

`30^o`

(B)

`45^o`

(C)

`60^o`

(D)

`90^o`

Solution:

Centre of circle `S_1` is `C_1 (3,- 2)`.

Centre of circle `S_2` is `C_2 (2, - 3)`

Let `r_1` and `r_2` be their radii

`r_1 = sqrt (9+ 4 -11) = sqrt 2`

and `r_2 = sqrt (4+ 9 -9 ) =2`

Also,

`C_1 C_2 = sqrt ( (3-2)^2 + (-2+3)^2) = sqrt 2`

Suppose, two circles intersect at P.
Then,

`cos angle C_1 P C_2 = ( C_1 P^2 + C_2 P^2 - C_1 C_(2)^2 )/( 2C_1 P * C_2 P)`

`= (2+ 4 -2 )/( 2 xx sqrt 2 xx 2 ) = 1/(sqrt 2)`

`=> cos angle L_1 PC_2 = cos 45^o`

`=> angle `C_1 P C_2 = 45^`o`
Correct Answer is `=>` (B) `45^o`
Q 2855167964

Consider the two circles

`S_1 = x^2 + y^2 - 6x + 4y + 11 = 0 `and

`S_2 = x^2 + y^2 - 4x + 6y + 9 = 0`.
The equation of common chord is

(A)

`y - x = 1`

(B)

`x- y = 1`

(C)

`x+ y -1 = 0`

(D)

`x+ y = 1`

Solution:

Equation of common chord is given by

`S_1 - S_2 = 0`

`=> - 6x+ 4y+ 11-(-4x + 6y + 9) = 0`

`=> - 2x - 2 y + 2 = 0 => x + y - 1 = 0`
Correct Answer is `=>` (C) `x+ y -1 = 0`
Q 2875067866

Find the number of common tangents drawn to
`S_1` and `S_2`

(A)

2

(B)

1

(C)

3

(D)

4

Solution:

Here `| r_1 - r_2 | < C_1 C_2 < | r_1 + r_2 |`

Clearly, circles cut each other at two
points. Hence, number of common
tangents will be 2.
Correct Answer is `=>` (B) 1
Q 2855867764

Find `C_1 C_2`.

(A)

2

(B)

`sqrt 29`

(C)

`sqrt 41`

(D)

`5`

Solution:

Here, `C_1 (0,0)` and `C_2 (4,5)`

`:. C_1 C_2 = sqrt (16 +25) = sqrt (41)`
Correct Answer is `=>` (D) `5`
Q 2825567461

Which of the following is true?

(A)

`r_1 > r_2`

(B)

`r_1 < r_2`

(C)

`r_1 = r_2`

(D)

`r_2 > r_1`

Solution:

From circle `S_1 , r_1 = sqrt (16) = 4`

From circle `S_2 , r_2 = sqrt (16 + 25 +23) = 8`

Clearly, `r_1 < r_2`
Correct Answer is `=>` (B) `r_1 < r_2`
Q 2865467365

A circle always passes
through the fixed points (a, 0) and (-a, 0).
If given points are the ends of diameter, then the
equation of circle is

(A)

`x^2 + y^2 = a^2`

(B)

`x^2 + y^2 + a^2 = 0`

(C)

`x^2 + y^2 + 2x + 2y = a^2`

(D)

`x^2 + y^2 -2x -2y = a^2`

Solution:

Since, (a, 0) and (-a, 0) arc the end
points of diameter, then equation of the
circle is

`(x - a) (x + a)+ (y- 0) (y-0)= 0`

`=> x^2 - a^2 + y^2 = 0 => x^2 + y^2 = a^2`
Correct Answer is `=>` (A) `x^2 + y^2 = a^2`
Q 2865467365

A circle always passes
through the fixed points (a, 0) and (-a, 0).
How many tangents can be drawn from the
origin to the circle that we get in above question?

(A)

0

(B)

1

(C)

2

(D)

3

Solution:

Since, origin lies inside the circle

`x^2 + y^2 = a^2`. Hence, no tangent can be
drawn.
Correct Answer is `=>` (A) 0
Q 2815367260

Consider the circle

`S ≡ x^2 + y^2 - 6x + 12y + 15 = 0`.
The equation of circle which is concentric with
circle S and has area double of its area is

(A)

`x^2 + y^2 - 6x + 12y -15 = 0`

(B)

`x^2 + y^2 -6x + 12y+ 40= 0`

(C)

`x^2 + y^2 - 6x + 12y+ 45= 0`

(D)

None ot the above

Solution:

Equation of circle concentric to given
circle is

`S_1 ≡ x^2 + y^2- 6x + 12 y + k = 0` ... (i)

As, area of circle `(S_1 )`

`= 2 `area of (given) circle (S)

`:.` Radius of circle `( S_1 ) = sqrt 2`

[radius of given circle]

`=> sqrt (9 + 36 -k) = sqrt (2) sqrt ( 9+ 36 -15 )`

`=> 45 -k = 60 => k = -15`

Hence, the required equation of circle is

`x^2 + y^2 - 6x + 12y - 15 = 0`
Correct Answer is `=>` (A) `x^2 + y^2 - 6x + 12y -15 = 0`
Q 2815367260

Consider the circle

`S ≡ x^2 + y^2 - 6x + 12y + 15 = 0`.
Which of the following line is a diameter of the
circle `S`?

(A)

`2y + 3x + 3 = 0`

(B)

`3y + 2x + 7 = 0`

(C)

`x + y - 3 = 0`

(D)

`x - y + 9 = 0`

Solution:

Centre of the circle S is `(3,- 6)`.

Since, centre `(3,- 6)` is satisfying only

`2y + 3x + 3 = 0`

`:. 2 y + 3x + 3 = 0` is the required line.
Correct Answer is `=>` (A) `2y + 3x + 3 = 0`
Q 2805167068

Consider the circle

`S ≡ x^2 + 4x + (y- 3)^2 = 0`.
From the point A(0, 3) on the circle a chord AB is
drawn and extended to a point P. Such that
AP = 2AB. The locus of P is

(A)

`x^2 + (y- 3)^2 = 0`

(B)

`x^2 + 4x + (y+3)^2 = 0`

(C)

`x^2 + 8x +(y-3)^2 =6`

(D)

`(x+4)^2 + (y- 3)^2 =16`

Solution:

Let the coordinates of P be (b, k).
Then, B is the mid-point of AP. So, the

coordinates of `B` are `(b/2 , ( k+3)/2)`

Since, B lies on the circle.

`x^2 + 4x + (y-3)^2 = 0`

`:. b^2/4 + 4 (b/2) + ( ( k+3)/2 -3 )^2 = 0`

`=> b^2 + 8b + (k-3)^2 = 0`

Hence, the locus of P(b, k) is

`x^2 + 8x +(y -3)^2 = 0`.
Correct Answer is `=>` (D) `(x+4)^2 + (y- 3)^2 =16`
Q 2805167068

Consider the circle

`S ≡ x^2 + 4x + (y- 3)^2 = 0`.
The distance of the origin from the centre of S is

(A)

`7/2`

(B)

`4`

(C)

`sqrt 13`

(D)

`sqrt 15`

Solution:

We have,

`S ≡ x^2 + 4x + y^2 + 9- 6y = 0`

or `S ≡ x^2 + y^2 + 4x- 6y + 9 = 0`

Centre of circle, `S = (- 2, 3)`

Required distance from origin

`= sqrt ( (-2)^2 + (3)^2 ) = sqrt (13)`
Correct Answer is `=>` (C) `sqrt 13`
Q 2885156967

Consider the following circles

I. `x^2 + y^2 + 4x- 6y- 12 = 0`

II. `x^2 + y^2 -12x-14y+60=0`

III . `x^2 + y^2 - 10 x + 8y + 18 = 0`

Which of the above circles has equal area?

(A)

I,II

(B)

I, III

(C)

II, III

(D)

I,II, III

Solution:

I. Centre = (- 2, 3)

Radius `= sqrt ( (-2)^2 + 3^2 + 12)`

`= sqrt (25) = 5`

`:.` Area ` = 25 pi`

II. Centre `= ( 6,7)`

Radius `= sqrt (6^2 + 7^2 -60) = sqrt 25 = 5`

`:.` Area `= 25 pi`

III. Centre `= ( 5, -4)`

Radius `= sqrt (5^2 + (-4)^2 -18 ) = sqrt 23`

`:.` Area `= 23 pi`

So, I and II have equal area .
Correct Answer is `=>` (A) I,II
Q 2835856762

Consider the following statements
I. The equation` x^2 + y^2 + 2x- 10 y+ 30 = 0`,

represents the equation of a circle.

II. If point (0, g) lies inside the circle
`x^2 + y^2 +2 gx + c = 0` , then `c< 0`.

Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. The radius of this circle

`= sqrt (1^2 + (-5)^2 -30 ) = sqrt ( (-4) )`

which is imaginary.
So, Statement I is false.

II. Since, (0, g) lies inside

`S ≡ x^2 + y^2 + 2gx + c = 0`

`:. S_1 = 0+ g^2 + 0 +c < 0`

`=> g^2 + c < 0 => c < 0` `[ ·: g^2 > 0]`

So, Statement II is true.
Correct Answer is `=>` (B) Only II
Q 2855756664

Consider the following statements
I. Number of circles touching the given three
non-concurrent lines is 4.
II. Number of circles passing through (1, 2), (4, 8)
and (0, 0) is one.
Which of the above statement(s) is I are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

None ot these

Solution:

Correct Answer is `=>` (A) Only I
Q 2835756662

A square is inscribed in a circle
`x^2 + y^2- 2x + 4y + 3 = 0`. Its sides are parallel to
the coordinate axes. Then, one of the vertex of
the square is

(A)

`(1+ sqrt 2 , -2 )`

(B)

`(1- sqrt 2 , -2 )`

(C)

`(1 , -2 + sqrt 2)`

(D)

None of the above

Solution:

Centre of the circle is (1, - 2) and
sides of the inscribed square arc parallel
to the coordinate axes. Hence, no vertex
of the square can have its coordinates `x`
as `1` and `y` as` - 2` .
Correct Answer is `=>` (D) None of the above
Q 2885656567

What is the equation to the circle which touches
both the axes and has centre on the line
x+y=4'?

(A)

`x^2 + y^2 - 4x + 4y + 4= 0`

(B)

`x^2 + y^2 - 4x- 4y + 4 = 0`

(C)

`x^2 + y^2 + 4x- 4y- 4 = 0`

(D)

`x^2 + y^2 + 4x + 4y- 4 = 0`

Solution:

We know that, the equation of circle,
which touches both the axes, is

`x^2 + y^2 - 2 r x- 2 r y + r^2 = 0` ... (i)

The centre ( r, r) of this circle lies on the

line `x + y = 4`.

`:. r+ r = 4 => r = 2`

On putting the value of r in Eq. (i), we

get `x^2 + y^2 - 4x - 4y + 4 = 0`

which is the required equation of the
circle.
Correct Answer is `=>` (B) `x^2 + y^2 - 4x- 4y + 4 = 0`
Q 2855656564

Consider a circle of radius R. What is the length
of a chord which subtends an angle `theta` at the centre'?

(A)

` 2 R sin ( theta/2)`

(B)

`2R sin theta`

(C)

`2 R tan ( theta/2)`

(D)

`2 R tan theta`

Solution:

In `Delta ADO`

`sin theta/2 = (AD)/(OA) = (AD)/R`

`=> AD=R sin theta/2`

`:.` Length of the chord

`AB = 2AD = 2R sin ( theta/2 )`
Correct Answer is `=>` (A) ` 2 R sin ( theta/2)`
Q 2815656560

The centre of the circle `(x- alpha)^2 + (y - beta )^2 = 9` lies
on the straight line `x = y` and the circle touches
the circle `x^2 + y^2 = 1 `externally. What are the
values of `alpha , beta`?

(A)

`alpha = pm 2 sqrt 2 , beta = pm 2 sqrt 2`

(B)

`alpha = pm sqrt 2 , beta = pm sqrt 2`

(C)

`alpha = 0, beta = 0`

(D)

`alpha = 2, beta = 2`

Solution:

The centre and radius of circle

`(x- alpha)^2 + (y -beta)^2 = 9` are `(alpha , beta)` and
3, respectively.

Since, `(alpha , beta )` lies on the straight line

`y = x`

`:. alpha = beta` .............(i)

Now, this circle touches the circle
`x^2 + y^2 = 1` externally.

`:. alpha^2 + beta^2 = 3 +1 => alpha^2 + beta^2 = 4`

`=> 2 al[ha^2 =4` [using Eq. (i)]

`=> alpha = pm sqrt 2`

`:. alpha = pm sqrt 2 ` and `beta = pm sqrt 2`
Correct Answer is `=>` (B) `alpha = pm sqrt 2 , beta = pm sqrt 2`
Q 2815556460

Under which one of the following conditions does
the circle `x^2 + y^2 + 2gx + 2fy + c = 0` meet the
X- axis in two points on opposite sides of the
origin?

(A)

`c > 0 `

(B)

`c < 0 `

(C)

`c = 0 `

(D)

`c le 0`

Solution:

The circle

`x^2 + y^2 + 2gx + 2fy + c = 0`,

meets X-axis (y = 0) in two points on
opposite sides of origin.

It means `x^2 + 2gx + r = 0`

`=> x = ( -2g pm sqrt (4g^2 - 4c) )/2`

`= -g pm sqrt (g^2 - c)`

Circle meets the X-axis in two points on
opposite side of origin
Hence,

`-g + sqrt (g^2 -c) > 0 , -g - sqrt (g^2 -c) < 0`

`=> sqrt (g^2 -c) > g => g^2 -c > g^2 => c < 0`
Correct Answer is `=>` (B) `c < 0 `
Q 2835456362

Locus of the centre of the circle which always
passes through the fixed points (a, 0) and (-a, 0) is

(A)

x=1

(B)

x+y =6

(C)

x+y =2a

(D)

x= 0

Solution:

If( b, k) is centre C and A, B be the
given points, then

`CA^2 = CB^2 => 4ab = 0 => b = 0`

`:. x = 0`
Correct Answer is `=>` (D) x= 0
Q 2885356267

Equation of a circle passing through ( -1, 2) and
concentric with the circle `x^2 + y^2- 3x + 4y - c = 0`
is

(A)

`x^2 + y^2 - 3x + 4y - 1 = 0`

(B)

`x^2 + y^2 - 3x + 4y = 0`

(C)

`x^2 + y^2 - 3x + 4y + 2 = 0`

(D)

None of these

Solution:

The terms of x and y will remain
the same as the two circles arc
concentric. The new constant` lambda = 0` as it
passes through `( -1,-2 )`. By putting
point `( -1, -2)` in the equation.
Correct Answer is `=>` (B) `x^2 + y^2 - 3x + 4y = 0`
Q 2815356260

The radius of the circle passing through the
point `P (6, 2)`, two of whose diameter are x + y = 6
and x + 2y = 4 is

(A)

`10 `

(B)

`2 sqrt 5`

(C)

`6`

(D)

`4`

Solution:

The centre of circle is the intersection
of equations of the diameter
`x + y = 6` ... (i)

and `x + 2y = 4 `... (ii)

On solving Eqs. (i) and (ii), we get

`x = 8`

and ` y = -2`

`:.` Radius of circle is the distance
between `P( 6, 2)` and `C(8, - 2)`

Radius `= sqrt ( (8-6)^2 + (2+2)^2 )`

`=2 sqrt 5`
Correct Answer is `=>` (B) `2 sqrt 5`
Q 2885256167

The lines 2x- 3y = 5 and 3x- 4y = 7 are
diameters of a circle of area 154 sq units. Then,
the equation of this circle is

(A)

`x^2 + y^2 + 2x - 2y = 62`

(B)

`x^2 + y^2 + 2x- 2y = 47`

(C)

`x^2 + y^2 -2x + 2y= 47`

(D)

`x^2 + y2 -2x + 2y = 62`

Solution:

If r is radius of the circle,

then `pi r^2 = 154`

`:. r^2 = 154 xx 7/22 = 49` [taing `pi = 22/7 `]

`=> r= 7`

Also, solving the equation of two given
diameters, we get the coordinates of the
centre as `(1, -1)`.

Hence, the equation of the circle is

`(x -1)^2 + (y + 1)^2 = 7^2 = 49`

`=> x^2 + y^2 - 2x + 2y = 47`
Correct Answer is `=>` (C) `x^2 + y^2 -2x + 2y= 47`
Q 2835256162

A square is inscribed in the circle `x^2 + y^2 + 2gx
+ 2fy + c = 0` of radius r, then length of its side is

(A)

`r`

(B)

`r sqrt 2`

(C)

`1/2 r`

(D)

`sqrt 2`

Solution:

If a is the side of the square inscribed
in a circle of radius r, then `a^2 + a^2 =
(text (Diameter))^2`

`:. 2a^2 = 4 r^2`

`=> a = r sqrt 2`
Correct Answer is `=>` (B) `r sqrt 2`
Q 2825256161

Equation of circle which passes through the
points (1, -2) and (3,- 4) and touch the X-axis is

(A)

`x^2 + y^2 + 6x + 2y + 9 = 0`

(B)

`x^2 + y^2 + 10 x + 20y + 25 = 0`

(C)

`x^2 + y^2 + 6x + 4y + 9 = 0`

(D)

None of the above

Solution:

Since, the circle touches X-axis.
`:. (x- b)^2 + (y- k)^2 = k^2` ... (i)

Also, it passes through given points

`(1- b)^2 + (-2- k)^2 = k^2` ... (ii)

and `(3-b)^ 2 + (-4-k )^2 =k^2` ... (iii)
On subtracting Eq. (iii) from Eq. (ii), we
get b = k + .5
On solving these equations, we get

`k = -10,-2 `and `b = -5 , 3`

By putting the values of
`(b, k) = (-5, -10)` or `(3, 2)` in Eq. (i),

we get `x^2 + y^2 + 10x + 20y + 25 = 0`
Correct Answer is `=>` (B) `x^2 + y^2 + 10 x + 20y + 25 = 0`
Q 2805156068

The equation of the circle passing through (4, 5)
having the centre at (2, 2) is

(A)

`x^2 + y^2 + 4x + 4y- 5 = 0`

(B)

`x^2 + y^2 - 4x- 4y- 5 = 0`

(C)

`x^2 + y^2 -4x = 13`

(D)

`x^2 + y^2 -4x-4y + 5 = 0`

Solution:

Here, `r=` Distance between ( 4, 5)
and (2, 2)

`:. r^2 = 4 + 9 = 13`

`=> (x-2)^2 + (y-2)^2 = 13`

`=> x^2 + y^2 -4x -4y - 5 = 0`
Correct Answer is `=>` (B) `x^2 + y^2 - 4x- 4y- 5 = 0`
Q 2835156062

For the equation `ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0`,
where `a ne 0`, to represent a cirele, the condition
will be

(A)

a = b and c = 0

(B)

f = g and h = 0

(C)

a=b band h = 0

(D)

f = g and c = 0

Solution:

The general equation of circle is

`x^2 + y^2 + 2gx + 2fx + c = 0` .... (i)

Hence, on comparing the given equation
with Eq. (i), we get

`a= b` and `b = 0`
Correct Answer is `=>` (C) a=b band h = 0

Advance Practice Problems For NDA

Q 2418656500

For the given circles `x^ 2 + y^ 2 6x - 2y + 1 = 0` and
`x^2+y^2 +2x- 8y+13=0`, which of the
following is true'?
BCECE Stage 1 2016
(A)

One circle lies inside the other

(B)

One circle lies completely outside the other

(C)

Two circle intersect in two points

(D)

They touch each other externally

Solution:

The centres of given circles are `C_1(3, 1)` and `C_2(- 1, 4)` and
corresponding radii are

`r_1=sqrt((3)^2+(1)^2-1) =3`

and `r_2=sqrt((-1)^2+(4)^2-13)=2`

Now, `C_1C_2=sqrt((-1-3)^2+(4-1)^2)=5`

`C_1C_2=r_1+r_2`

Hence, two Circles touch externally.
Correct Answer is `=>` (D) They touch each other externally
Q 2488167907

A ladder `10 m` long rests against a vertical
wall with the lower end on the horizontal
ground. The lower end of the ladder is
pulled along the ground away from the
wall at the rate of `3 cm//s`. The height of the
upper end while it is descending at the rate
of `4 cm//s` is
BCECE Stage 1 2016
(A)

`4 sqrt 3` m

(B)

`5 sqrt 3` m

(C)

`5 sqrt 2` m

(D)

`6` m

Solution:

Let `AB = x, BC = y` and `AC =10` m

`:.x^2+y^2=100`.................(i)

`=> 2x(dx)/(dt)+2y(dy)/(dt)=0`

`=> 2x(3)- 2y(4) = 0`

`[ ∵ (dx)/(dt)=3 m//s ,(dy)/(dt)=-4m//s ]`


On putting this value in Eq (i), we get


`16/9 y^2+y^2=100=> y=6 m`
Correct Answer is `=>` (C) `5 sqrt 2` m
Q 2579791616

A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve `x^2 + y^2 = 4` with `x + y = a`. The set containing the value of a is
BCECE Mains 2015
(A)

`{-2,2}`

(B)

`{-3,3}`

(C)

`{-4,4}`

(D)

`{-5,5}`

Solution:

To make the given curves `x^2 + y^2 = 4` and

`x + y = 9` homogeneous,

`x^2 + y^2 - 4((x + y)/a)^2 = 0`

`=> a^2 (x^2 + y^2) - 4(x^2 + y^2 + 2xy) = 0`

`=> x^2 (x^2 - 4) + y^2 - 4) - 8xy = 0`

Since, this is perpendicular pair of straight lines.

`:. a^2 - 4 + a^2 = 0`

`=> a^2 = 4`

`=> a = ±2`

Hence, required set of a is `{- 2, 2}`.
Correct Answer is `=>` (A) `{-2,2}`
Q 2509791618

The radius of any circle touching the lines `3x - 4y + 5 = 0` and `6x - 8y - 9 = 0` is
BCECE Mains 2015
(A)

`1.9`

(B)

`0.95`

(C)

`2.9`

(D)

`1.45`

Solution:

Given lines `3x- 4y + 5 = 0` and `3x - 4y - 9/2 = 0`,

which are parallel to each others.

`:.` Perpendicular distance,

` d = | ( 5 + 9/2)/sqrt(3^2 + 4^2) | = (19)/(10)`

`:.` Radius of circle `= d/2 = (19)/(20) = 0.95`
Correct Answer is `=>` (B) `0.95`
Q 2579891716

A common tangent to `9 x^2 - 16y^2 = 144` and `x^2 + y^2 = 9`, is


BCECE Mains 2015
(A)

` y = 3/sqrt7 x + (15)/sqrt7`

(B)

` y = 3 sqrt(2/7) x + (15)/sqrt7`

(C)

` y = 2 sqrt(3/7)x + 15 sqrt(17)`

(D)

None of these

Solution:

Given, curves are `9x^2 - 16y^2 = 144`

and `x^2 + y^2 = 9`

Let the equation of common tangent be

`y = mx + c`

Since, `y = mx + c` is a tangent to `x^2/(16) - y^2/9 = 1`.

`:. c^2 = 16m^2 - 9 [ ∵ c^2 = a^2 m^2 - b^2 ]`

Similarly, `y = mx + c` is a tangent to `x^2 + y^ 2 = 9`.

`:. c = 3 sqrt(m^2 +1) => c^2 = 9 (1 + m^2)`......(ii)

From Eqs. (i) and (ii), we get

` 16m^2 - 9 = 9 + 9m^2 => m^2 = (18)/7 => m = 3 sqrt(2/7)`

From Eq. (ii), we get

` c^2 = 9 (1 + (18)/7) => c^2 = 9 ( (25)/7)`

` => c = (pm 15)/sqrt7`

Hence, ` y = 3 sqrt(2/7) x + (15)/sqrt7`
Correct Answer is `=>` (B) ` y = 3 sqrt(2/7) x + (15)/sqrt7`
Q 2428780601

Consider the circles `x^2 + (y - 1)^2 = 9,`
`(x - 1)^2 + y^2 = 25`. They are such that
BCECE Stage 1 2015
(A)

these circles touch each other

(B)

one of these circles lies entirely inside the other

(C)

each of these circles lies outside the other

(D)

they intersect in two points

Solution:

Centres and radii of the given circles are
Centres `C_1 (1, 0) C_2 (1, 0)`
Radii `r_1 = 3 , r_2 = 5`
clearly `C_1C_2 = sqrt2 < r_2 - r_1`
Therefore, one circle lies entirely inside the other
Q 2500601518

The circles `x^2 +y^2 -5x+6y+15=0` and `x^2 + y^2 - 2x + 6y + 6 = 0` touch each other
BCECE Stage 1 2014
(A)

internally

(B)

externally

(C)

Do not say anything

(D)

None of these

Solution:

The centre and radii of given circles are

`C_1 (1, -3), C_2 (5/2, -3 )` and `r_1 = sqrt (1 + 9 -6) =2`

`r_2 = sqrt(25/4 + 9 -15 ) = sqrt((25 -24) /4) =1/2`

Now, `C_1 C_2 = sqrt((1 -5/2)^2 + (-3 + 3)^2)`

`= sqrt((3/2)^2 + 0) =3/2`

and `r_1 -r_2 =2 -1/2 =3/2`

`because r_1 -r_2 = C_1 C_2`

Hence , the two circles touch each other externally
Correct Answer is `=>` (B) externally
Q 2510845719

The line `(x - 2) cos theta + (y - 2) sin theta = 1` touches a circle for all value of `theta`, then the equation of circle is
BCECE Stage 1 2013
(A)

`x^2 + y^2 - 4x - 4y + 7 = 0`

(B)

`x^2 + y^2 + 4x + 4y + 7 = 0`

(C)

`x^2 + y^2 - 4x - 4y - 7 = 0`

(D)

None of the above

Solution:

Given line is

`(x - 2) cos theta + (y - 2)sin theta = 1`

` = cos^2 theta + sin^2 theta `

On comparing we get,

` x - 2 = cos theta ` ... (i)

` y - 2 = sin theta ` ... (ii)

and On squaring and then adding Eqs. (i) and

(ii), we get

`(x- 2)^2 + (y -2)^2 = cos^2 theta + sin^2 theta`

`=> (x-2)^2 + (y-2)^2 = 1`

`=> x^2 + y^2 - 4x - 4y + 7 = 0`
Correct Answer is `=>` (A) `x^2 + y^2 - 4x - 4y + 7 = 0`
Q 2520145911

If `(-3, 2)` lies on the circle `x^2 + y^2 + 2gx +2fy + c = 0` which is concentric with the circle `x^2 + y^2 + 6x + 8y - 5 = 0`, then `C` is equal to
BCECE Stage 1 2013
(A)

`11`

(B)

`-11`

(C)

`24`

(D)

`100`

Solution:

Equation of family of concentric circles to the

circle `x^2 + y^2 + 6x + 8y - 5 = 0`

`x^2 + y^2 + 6x + 8y + lamda = 0`

which is similar to `x^2 + y^2 + 2gx +2fy + c = 0`.

Thus, the point `(-3, 2)` lies on the circle.

`x^2 + y^2 + 6x + 8y+ c = 0`

`=> (-3)^2 + (2)^2 + 6(-3) + 8(2)+ c = 0`

` => 9 + 4 - 18 + 16 + c = 0`

` :. c = -11`
Correct Answer is `=>` (B) `-11`
Q 2459480314

The other end of the diameter through the point `( -1, 1)` on the circle `x^2 + y^2 - 6x + 4y - 12 = 0` is
BCECE Stage 1 2012
(A)

`(-7, 5)`

(B)

`(-7,- 5)`

(C)

`(7, - 5)`

(D)

`(7, 5)`

Solution:

Given, circle is `x^2 + y^2- 6x + 4y -12 = 0`
Centre of this circle is `(3,- 2)`
Let other end of the diameter is `(alpha,beta)`).



`therefore (alpha-1)/2 = 3 , (beta+1)/2 = -2`


`=> alpha = 7 , beta = -5`


`therefore` Other end of the diameter is `(7, - 5)`.
Correct Answer is `=>` (C) `(7, - 5)`
Q 2419580410

The number of common tangents to the circles `x^2 + y^2 = 4` and `x^2 + y^2 - 6x- 8y + 24 = 0` is
BCECE Stage 1 2012
(A)

`3`

(B)

`4`

(C)

`2`

(D)

`1`

Solution:

The centres and radii of circles `C_1 (0, 0), C_2 (3, 4)` and


`r_1 = 2 , r_2 = sqrt(9+16-24) = 1`


Now `C_1C_2 = sqrt((3-0)^2+(4-0)^2) = 5`


`r_1+r_2 = 2+1 = 3`



since `C_1C_2 > r_1+r_2`


`therefore` Number of common tangents = 4.
Correct Answer is `=>` (B) `4`
Q 2409745618

The circle `S_1` with centre `C_1 (a_1, b_1)` and radius touches externally the circle `S_2` with centre `C_2 (a_2, b_2 )` and radius `r_2`. If the tangent at their common point passes through the origin, then
BCECE Stage 1 2011
(A)

`(a_1^2+ a_2^2) + (b_1^2 + b_2^2) = r_1^2+r_2^2`

(B)

`(a_1^2-a_2^2)+(b_1^2-b_2^2)=r_1^2-r_1^2`

(C)

`(a_1^2-b_2)^2+(a_2^2+b_2^2)=r_1^2+r_2^2`

(D)

`(a_1^2-b_1^2)+(a_1^2+b_2^2)=r_1+r_2^2`

Solution:

Given two circles are

`S_1 equiv (x- a_1)^2+ (y- b_1^2)=r_1^2`...............(i)

and `S_1 equiv (x-a_2)^2+(y-b_2^2)=r_2^2`...........(ii)

The equation of the common tangent of these two circles is given by `S_1 -S_2 = 0` i.e.,

`2x (a_1-a _2 )+ 2y (b_1 - b_2 ) + (a_2^2 + b_2^2 )- (a_1^2+b_1^2)+r_1^2-r_2^2=0`

If this passes through the origin, then

`(a_2^2+b_2^2)-(a_1^2+b_1^2)+r_1^2-r_2^2=0`

`=>(a_2^2-a_1^2)+(b_2^2-b_1^2)=r_2^2-r_1^2`

`=>(a_1^2-a_2^2)+(b_1^2-b_2^2)=(r_1^2-r_2^2)`
Correct Answer is `=>` (B) `(a_1^2-a_2^2)+(b_1^2-b_2^2)=r_1^2-r_1^2`
Q 2419445319

The equation of a line parallel to the line
`3x+ 4y = 0` and touching the circle
`x^2 + y^2 = 9` in the first quadrant, is
WBJEE 2016
(A)

`3x+4y=15`

(B)

`3x+4y=45`

(C)

`3x +4y =9`

(D)

`3x +4y =27`

Solution:

Equation of circle is `x^2 + y^2 = 9`.............(i)

And equation of line is `3x + 4y = 0`...................(ii)

Equation of line parallel to the line (ii),

`3x + 4y = k`

`=> y = (-3x)/4 +k/4`..............(iii)

`:.` Circle touches the line (iii), so by the condition of
tangency,

`c^2 = a^2 (1+m^2)`

`=> (k/4)^2 = 9 [1+ (-3/4)^2]`

`[ :. c= k/4, a=3 ,m =-3/4]`

`=> k= pm 15`

Hence, `3x + 4y = 15` touches the circle in the first
q~adrant.
Correct Answer is `=>` (A) `3x+4y=15`
Q 2419645519

The locus of the mid-points of chords of the
circle `x^2 + y^2 = 1`, which subtends a right
angle at the origin, is
WBJEE 2016
(A)

`x^2 + y^2 = 1/4`

(B)

`x^2 + y^2 =1/2`

(C)

`xy =0`

(D)

`x^2 - y^2 =0`

Solution:

Let `(h ,k)` be the coordinates of the mid-point of a chord,

which subtends a right angle at the origin.

Then, equation of the chord is

`hx + ky- 1 = h^2 +k^2 - 1` [using `T = S' `]

`=> hx + ky = h^2 + k^2`

the combined equation of the pair of lines joining the
origin to the points of intersection of `x^2 + y^2 = 1` and

`hx + ky = h^2 + k^2` is

`x^2 + y^2 -1 ((hx +ky)/(h^2 + k^2))^2 =0`

Lines given by the above equation are at right angle.

Therefore, coefficient of `x^2 +` coefficient of `y^2 = 0`

i.e., `h^2 + k^2 =1/2`

`:. x^2 + y^2 = 1/2`
Correct Answer is `=>` (B) `x^2 + y^2 =1/2`
Q 2523645541

The angle of intersection between the curves `y=[|sin x|+| cos x|]` and `x^2+y^2=10` where `[x]` denotes the greatest integer `le x`, is
WBJEE 2014
(A)

`tan^(-1) 3`

(B)

`tan^(-1) (3)`

(C)

`tan^(-1) sqrt 3`

(D)

`tan^(-1) (1//sqrt 3)`

Solution:

Given, `y = [| sin x| +| cos x| ]` and `x ^2 + y^2 = 10`

We know that `(| sin x | + | cos x | ) in(1, sqrt 2)`

`:. y=1`

The point of intersection of given curve is

`x^2+1^2=10`

`=> x^2=9`

`=> x= pm 3`

`:.` :. Point of intersection is `(±3, 1)`

Now, `x^2+y^2=10`

`=> 2x+2y (dy)/(dx)=0`

`=> (dy)/(dx)=x/y`

At point `( -3, 1)`

`(dy)/(dx) =3/1=3 => m_1=3`

slope of line `y =1` is `m_2 =0`

`:.` Angle between two curves is `tan theta=(m_1-m_2)/(1+m_1m_2)=3`

`=> theta=tan^(-1) (3)`
Correct Answer is `=>` (A) `tan^(-1) 3`
Q 2513256140

If the circle `x^2 + y^2 + 2gx + 2fy + c = 0` cuts the three circles `x^2 + y^2 -5 = 0`, `x^2+y^2-8x-6y+ 10=0` and `x^2 + y^2- 4x + 2y- 2 = 0` at the extremities of their diameters, then
WBJEE 2014
(A)

`c=-5`

(B)

`fg=147//25`

(C)

`g + 2f= c + 2`

(D)

`4f = 3g`

Solution:

Centres and constant terms in the circles

`x^2+ y^2 - 5 = 0, x^2 + y^2 - 8x - 6y + 1 0 = 0` and

`x ^2 +y^2-4x+2y-2=0` are `C_1{(0,0), c_1 =-5`,

`C_2 (4, 3), c_2 = 10` and `C_3(2, -1), c_3 = -2`

Also, centre and constant of circle

`x^2 + y^2 + 2gx + 2fy + c = 0`

is `C_4 ( -g, -f)` and `c_4 = c`

Since, the first circle intersect all the three at the extremities of diameter, therefore they are orthogonal to each other.

`:. 2(g_1g_2+f_1f_2)=c_1+c`

`:. 2 [-g(0) +(-f) (0)] =c =5`

`=> c=5`

`2 [-g(4) +(-f) (3)] = c + 10`

`=> -2(4g+3f)=5+10`

`=> 4g+3f=-15/2`...............(i)

and `2 [-g(2) +(-f) (-1)] = c -2`

`=> -2g+f=3/2`...............(iii)

On solving Eqs. (i) and (ii), we get

`f=-9/10` and `g=-12/10`

`:. fg=(-9)/10 xx -12/10 =27/25`

and `4f=4 xx (-9)/10`

`=> 4f=(-36)/10`

`=> 4f=3g`
Correct Answer is `=>` (D) `4f = 3g`
Q 2581545427

A point `P` lies on the circle `x^2 + y^2 = 169`. If `Q = (5, 12)` and `R = ( -12, 5)`, then the `angleQPR` is
WBJEE 2013
(A)

`pi/6`

(B)

`pi/4`

(C)

`pi/3`

(D)

`pi/2`

Solution:

Given equation of circle is

`x^2 + y^2 = 169`


Its centre `= ( 0, 0)` and radius `= 13`


Now, slope of `OR = (-5)/12 = m_1` `(because text(slope) = (y_2-y)/(x_2-x))`


and slope of `OQ = 12/5 = m_2`


`because m_1m_2 = -1 => anglePOQ = pi/2`

We know that, angle made by the chord of circle at circumference is equal to the half of the angle
made by the same chord at the centre of circle

`therefore angleQPR = 1/2 . angleROQ = 1/2xxpi/2 = pi/4`
Correct Answer is `=>` (B) `pi/4`
Q 2501745628

A point moves, so that the sum of squares of its distance from the points `(1, 2)` and `( -2, 1)` is always 6. Then, its locus is
WBJEE 2013
(A)

the straight line `y -3/2 = -3(x+1/2)`

(B)

a circle with centre `(-1/2 , 3/2)` and radius `1/sqrt2`

(C)

a parabola with focus `(1, 2)` and directrix passing through `(-2, 1)`

(D)

an ellipse with foci `(1, 2)` and `(-2, 1)`

Solution:

Let `P` be any point, whose coordinate is `(h, k)`
Given,

`P` moves, so that the sum of squares of its distances from the points `A(1,2)` and `B (-2, 1)` is 6

`(PA)^2+(PB)^2 = 6`


`=> (h-1)^2+(k-2)^2+(h+2)^2+(k-1)^2 = 6`


`=> h^2+1-2h+k^2+4-4k+h^2+4+4h+k^2+1-2k = 6`


`=> 2h^2+2k^2+2h-6k+4 = 0`


`=> h^2+k^2+h-3k+2 = 0`


`therefore` Required locus is


`x^2+y^2+x-3y+2 = 0` Which represent a circle


Whose centre is `(-1/2 , 3/2)` and radius ` = sqrt(1/4+9/4-2) = sqrt(5/2-2) =1/sqrt2`
Correct Answer is `=>` (B) a circle with centre `(-1/2 , 3/2)` and radius `1/sqrt2`
Q 2521045821

A circle passing through `(0, 0), (2, 6), (6, 2)` cut the x-axis at the point `P ne (0, 0)`. Then, the lenght of `OP`, where `O` is the origin, is
WBJEE 2013
(A)

`5/2`

(B)

`5/sqrt2`

(C)

`5`

(D)

`10`

Solution:

Let the equation of circle is

`x^2+y^2+2gx+2fy+C = 0` ...........(i)

When, circle (i) passes through the origin

then `C = 0` .............(ii)

When, circle (i) passes through the point `(2, 6)`.

then `4 + 36 + 4g + 12f + 0 = 0`
`=> 4g + 12f + 40 = 0`
`=> g+3f=-10` ............(iii)

When, circle (i) passes through the point `(6, 2)`
Then, `36 + 4 + 12g + 4f + 0 = 0` [from Eq. (i)]

`=> 12g + 4 f + 40 = 0`
`=> 3g+f+10=0`.............(iv)
On solving Eqs. (iii) and (iv), we get


`g = (-5)/2` and `f = (-5)/2`

`therefore` Equation of circle becomes

`x^2+y^2-5x-5y = 0` .........(v)


Circle cut the x-axis.
So, put `y = 0` in Eq. (v), we get


`x^2-5x = 0 => x(x-5) = 0`


`=> x = 5`


So, the circle cut the x-axis at point `P(5, 0)`.
`therefore` The length of `OP = 5`
Correct Answer is `=>` (C) `5`
Q 2541778623

The equation `2x^2+5xy-12y^2 = 0` represents a
WBJEE 2013
(A)

circle

(B)

pair of non-perpendicular intersecting straight lines

(C)

pair of perpendicular straight lines

(D)

hyperbola

Solution:

Given equation is

`2x^2 +5xy-12y^2 = 0` ... (i)

`2x^2 + 8xy- 3xy -12y^2 = 0`


`2x(x + 4y)- 3y(x + 4y) = 0`

`(x + 4y) (2x - 3y) = 0`

`x + 4y = 0` and `2x - 3y = 0`

which represent a pair of straight lines.

Compair Eq. (i) with `ax^2 + 2hxy + by^2 = 0`

We get `a = 2` and `b = -12`
`because a+b ne 0`


So, lines are not perpendicular to each other.
Hence, it is a pair of non-perpendicular intersecting straight lines.
Correct Answer is `=>` (B) pair of non-perpendicular intersecting straight lines
Q 2511078820

If one end of a diameter of the circle `ax^2 +3y^2-9x+6y+5=0` is `(1,2)`, then the other end is
WBJEE 2013
(A)

`(2, 1)`

(B)

`(2, 4)`

(C)

`(2, - 4)`

(D)

`(-4, 2)`

Solution:

Given equaitson of circle is
`3x^2 + 3y^2- 9x + 6y + 5 = 0`

`=> x^2+y^2-3x+2y +5/3 =0`

Centre `(3/2 , -1)` and radius `= sqrt(9/4+1-5/3) = sqrt(19/12) = 1/2sqrt(19/3)`

We know that. centre of the circle is the mid-point of the diameter.
Lives one and of point of diameter in (1, 2) Let the other end point of diameter is (h, k).
Then `(3/2 , -1) = ((1+h)/2 , (2+k)/2)`


`=> (1+h)/2 = 3/2`

`=> 1+h = 3`

`=> h = 2`

and `(2+k)/2 = -1`


`=> 2+k = -2`

`=> k = -4`
So, the other end point is (2, - 4).
Correct Answer is `=>` (C) `(2, - 4)`
Q 2512134930

The equations of the circles, which touch both the axes and the line `4x + 3 y = 12` and have centres in the first quadrant, are
WBJEE 2013

(This question may have multiple correct answers)

(A) `x^2 + y^2 + x- y + 1 = 0`
(B) `x^2 + y^2 - 2x- 2y + 1 = 0`
(C) `x^2 + y^2 - 12x - 12y + 36 = 0`
(D) `x^2 + y^2 - 6x- 6y + 36 = 0`
Solution:

Radius `(r) =>` perpendicular distance on line `4x+3y = 12` from centre



`=> r = (|4r+3r-12|)/sqrt(16+9)`


`=> |7r-12| = 5r`


`=> 7r-12 = pm5r`


`therefore 2r = 12 => r = 6`


`12r = 12 => r = 1`


(i) When centre is (1, 1) and radius is 1, then equation of circle is



`(x-1)^2+(y-1)^2 = 1`


`=> x^2+y^2-2x-2y+1 = 0`

(ii) When centre is (6, 6) and radius is 2, then equation of circle is

`(x-6)^2+(y-6)^2 = 36`


`=> x^2+y^2-12x-12y+36 = 0`
Correct Answer is `=>` (B)
Q 2531578422

If the circles `x^2 + y^2 + 2x + 2ky + 6 = 0` and `x^2 + y^2 + 2ky + k = 0` intersect orthogonally then `k` is equal to
WBJEE 2012
(A)

` 2` or ` - 3/2`

(B)

` -2` or ` - 3/2`

(C)

` 2` or ` 3/2`

(D)

`- 2` or ` 3/2`

Solution:

Two circles are orthogonally if and only if

` 2(g_1g_2 + f_1f_2) = c_1 + c_2`

`=> 2 [(1 xx 0 + (k) k] = 6 + k`

` => 2k^2 = 6 + k`

`=> 2k^2 - k - 6 = 0`

` => 2k^2 - 4k + 3k - 6 = 0`

` => 2k (k - 2) + 3 (k - 2) = 0`

` => (k - 2)(2k + 3) = 0`

` => k = 2 , - 3/2`
Correct Answer is `=>` (A) ` 2` or ` - 3/2`
Q 2531678522

If four distinct points `(2k, 3k), (2, 0),(0, 3), (0,0)` lie on a circle, then
WBJEE 2012
(A)

`k < 0`

(B)

`0 < k < 1`

(C)

`k = 1`

(D)

`k > 1`

Solution:

Since, join of `(2, 0)` and `(0, 3)` subtends `90°` at

`(0, 0)`.

`=>` It is a diameter.

`:.` Equation is

`(x- 2)(x- 0) + (y- 0) (y- 3) = 0`

`x^2 + y^2 - 2x - 3y = 0`

`(2k, 3k)` lies on it

`=> 4k^2 + 9k^2 - 4k - 9k = 0`

`=> 13k^2 = 13k`

`=> k = 1`

Since, `k != 0` otherwise `(2k, 3k)` will be `(0, 0)`.
Correct Answer is `=>` (C) `k = 1`
Q 2512223130

The line `x = 2y` intersects the ellipse `x^2/4 + y^2 = 1` at the points P and Q. The equation of the circle with PQ as diameter is
WBJEE 2012
(A)

`x^2 + y^2 = 1/2`

(B)

`x^2 + y^2 = 1`

(C)

`x^2 + y^2 = 2`

(D)

`x^2 + y^2 = 5/2`

Solution:

Solving `x = 2y` ... (i)

and `x^2/4 + y^2 = 1` ... (ii)

Put `x = 2y` in Eq. (ii), we get

`(2y)^2/4 + y^2 = 1 => (4y^2)/4 + y^2 = 1`

`=> 2y^2 = 1 => y = ± 1/sqrt2`

`:.` From Eq. (i), `x = ± sqrt2`

`:. P ( sqrt2, 1/sqrt2)` and `Q (- sqrt2, - 1/sqrt2)`

`:.` Equation of circle with PQ as diameter is

`(x - sqrt2) (x + sqrt2) + (y - 1/sqrt2) (y + 1/sqrt2)`

`=> x^2 - 2 + y^2 - 1/2 = 0 => x^2 + y^2 = 5/2`
Correct Answer is `=>` (D) `x^2 + y^2 = 5/2`
Q 2532334232

Let `C_1` and `C_2` denote the centres of the circles `x^2 + y^2 = 4` and `(x - 2)^2 + y^2 = 1` respectively and let P and Q be their points of intersection. Then, the areas of `Delta C_1 PQ` and `C_2 PQ` are in the ratio
WBJEE 2012
(A)

`3:1`

(B)

`5 : 1`

(C)

`7 : 1`

(D)

`9 : 1`

Solution:

`x^2 + y^2 = 4` ....(i)

` (x - 2)^2 + y^2 = 1` ......(ii)

`C_1 = (0,0), C_2 = (2, 0)`

On solving Eqs. (i) and (ii), we get

`x = 7/4 , y = ± sqrt(15)/4`

`C_1N = 7/4 , PQ = (2sqrt(15))/4 = sqrt(15)/2`

`text (Area Delta C1PQ)/text(Area Delta C2PQ) = (1/2·PQ ·C_1N)/(1/2·PQ ·C_2N)`

`= (C_1 N)/(C_2N) = (7/4)/(2 - 7/4) = (7/4)/(1/4) = 7 : 1`

Hence, the required option is (c).
Correct Answer is `=>` (C) `7 : 1`
Q 1615891760

If a chord of the circle `x^2 + y^2 = 32` make equal
intercepts of length `l` on the coordinate axes, then
BITSAT 2016
(A)

`l in (0,8)`

(B)

`l in (-8,0)`

(C)

`l in (-8,8)`

(D)

`l in (-4 sqrt2, 4sqrt2)`

Solution:

The equation of lines making equal intercepts of length `I` on
the coordinate axes is

` x ± y = y = ± l`

This will be a chord of the circle `x^2 + y^2 = 32,` if lennth of the
perpendicular from the centre is less than radius.

`i.e. |(±l)/(sqrt(1+l)| < 4 sqrt2`

`=> |l| < 8`

`=> l in (-8,8)`
Correct Answer is `=>` (C) `l in (-8,8)`
Q 1542291133

The length of the `y`-intercept made by the circle `x^2 + y^2 - 4x -6y - 5 = 0` is
BITSAT 2013
(A)

`6`

(B)

`\sqrt{14}`

(C)

`2\sqrt{14}`

(D)

`3`

Solution:

To know the points on y axis, where the circle above intersects, we substitute `x` as `0`.

We thus get `y^2 - 6y - 5 = 0`

`\Rightarrow y = \ frac{6 \pm \sqrt{36 + 20}}{2} = \ frac{6 \pm \sqrt{56}}{2}`

`\Rightarrow y = 3 \pm \sqrt{14}`

So, the circle intersects the `y` axis at points `(0,3 - \sqrt{14})` and `(0,3 + \sqrt{14})`.

Thus, `y` intercept becomes `3 + \sqrt{14} - 3 + \sqrt{14} = 2\sqrt{14}`
Correct Answer is `=>` (C) `2\sqrt{14}`
Q 1512591430

If a circle passes through the point `(3, 4)` and cuts `x^2 + y^2 = 9` orthogonally, then the locus of its centre is `3x + 4y =\lambda`. Then `\lambda=`
BITSAT 2013
(A)

`11`

(B)

`13`

(C)

`17`

(D)

`23`

Solution:

Two circles are orthogonal, if the angle between them is `90^o.`

The condition which represents this is `2g_1g_2 + 2f_1f_2 = c_1 + c_2` for two circles `x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0` and `x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0`

Here, let the unknown circle be `x^2 + y^2 + 2gx + 2fy + c = 0`

The orthogonality condition thus becomes `0 + 0 = -9 + c`, implying `c` to be `9`.

Since the circle passes through `(3,4)`, we substitute that point in the circle to get `9 + 16 + 6g + 8f + 9 = 0`

i.e. `6g + 8f + 34 = 0`

`3g + 4f + 17 = 0`

Since the centre of the circle is `(-g,-f), x= -g` and `y = -f`

`\Rightarrow` locus is `-3x - 4y + 17 = 0`

i.e. `3x + 4y = 17`
Correct Answer is `=>` (C) `17`
Q 1533780642

The length of the tangent from `(5, 1)` to the circle `x^2+y^2+6x-4y-3=0` is:
BITSAT 2012
(A)

`7`

(B)

`49`

(C)

`63`

(D)

`21`

Solution:

`S_1: x^2+y^2+6x-4y-3=0` Center of the above circle, `S_1`, is `C(-3,2)` and radius `r=\sqrt{9+4+3}=4`

Let given point be `P(5,1)`

Now,

`CP =\sqrt{64+1}=\sqrt{65}`

Hence, length of required tangent is

`=\sqrt{CP^2-r^2}=\sqrt{65-16}=\sqrt{49}=7`
Correct Answer is `=>` (A) `7`
Q 1511745620

The length of tangent from point `(5, 1)` to
the circle `x^2 + y^2 + 6x - 4y - 3 = 0` is
BITSAT 2011
(A)

`81`

(B)

`29`

(C)

`7`

(D)

`21`

Solution:

Length of tangent `= sqrtS_1`

` = sqrt (5^2 + 1^2 + 30 -4-3)`

` = sqrt(49) = 7`
Correct Answer is `=>` (C) `7`
Q 1581345227

The equation of the normal to the circle

`x^2 + y^2 = a^2` at point `(x', y')` will be
BITSAT 2011
(A)

`x' y - xy' - 0`

(B)

`x x' - yy' = 0`

(C)

`x' y + xy' = 0`

(D)

`x x' + yy' = 0`

Solution:

Given, `x^2 + y^2 = a^2`

On differentiating w.r.t. `x`, we get

`2x+2y(dy)/(dx) = 0`

` => (dy)/(dx) = - x/y`

` => ((dy)/(dx))_(x',y')=-(x')/(y')`

`:. ` Equation of normal is

`y - y' = (y')/(x') (x - x')`

` => x' y - y ' x' = xy' - y' x'`

` => x' y - xy' = 0`
Correct Answer is `=>` (A) `x' y - xy' - 0`
Q 1500578418

The locus of centre of a circle which passes through the origin and cuts off a length of `4` unit from the line `x = 3` is
BITSAT 2009
(A)

`y^2 + 6x = 0`

(B)

`y^2 + 6x = 13`

(C)

`y^2 + 6x = 10`

(D)

`y^2 + 6y = 13`

Solution:

Let centre of circle be `C(-g, - f)`, then equation of circle passing
through origin be

`x^2 + y^2 + 2, gx + 2fy = 0`

`:. ` Distance, `d =|– g – 3| = g + 3`

In `Delta ABC, (BC)^2 = AC^2 + BA^2`

`=> g^2 + f^2 = (g + 3)^2 + 2^2`

`=> g^2 + f^2 = g^2 + 6g + 9 + 4`

`=> f^2 = 6g + 13`

Hence, required locus is `y^2 + 6x = 13`
Correct Answer is `=>` (B) `y^2 + 6x = 13`
Q 1520678511

The equation of the circle which passes through the origin and cuts orthogonally each of the circles `x^2 + y^2 - 6x + 8 = 0` and `x^2 + y^2 - 2x - 2y = 7` is
BITSAT 2009
(A)

`3x^2 + 3y^2 - 8x - 13y = 0`

(B)

`3x^2 + 3y^2 - 8x + 29y = 0`

(C)

`3x^2 + 3y^2 + 8x + 29y = 0`

(D)

`3x^2 + 3y^2 - 8x - 29y = 0`

Solution:

Let the required equation of circle be `x^2 + y^2 + 2gx + 2fy = 0`.

Since, the above circle cuts the given circles orthogonally.

`:. 2 (-3g) + 2f(0) = 8`

` => 2g = -8/3`

And `-2g - 2f = -7`

`=> 2f = +7 + 8/3 = 29/3`

Required equation of circle is

`x^2 + y^2 - 8/3 x + 29/3 y =0`

Or `3x^2 + 3y^2 - 8x + 29y = 0`
Correct Answer is `=>` (B) `3x^2 + 3y^2 - 8x + 29y = 0`
Q 1550078814

The inverse of the point `(1, 2)` with respect to the circle `x^2 + y^2 - 4x - 6y + 9 =0`, is
BITSAT 2008
(A)

`(2, 1)`

(B)

`(1, 1)`

(C)

`(0, 1)`

(D)

None of these

Solution:

the equation of w.r.
The equation of pole w.r.t. the point `(1, 2)` to the circle `x^2 + y^2 – 4x – 6y + 9 = 0` is
.
`x + 2y –2(x + 1) – 3(y + 2) + 9 = 0`

`x + y – 1= 0`

Since, the inverse of the point `(1, 2)` is the foot (alpha , beta ) of the perpendicular from the point `(1, 2)` to the line `x + y – 1`.

`(alpha-1)/1 = (beta-2)/2 = (1.1+1.2+1)/(1^2+1^2)`

`alpha-1=beta-2=-1`

`alpha=0`, `beta=1`

Hence required point is `(0, 1)`.
Correct Answer is `=>` (C) `(0, 1)`
Q 1510734610

The two circles `x^2 + y^2- 2x + 22y + 5 = 0` and `x^2 + y^2 + 14x + 6y + k = 0` intersect orthogonally provided `k` is equal to :
BITSAT 2006
(A)

`47`

(B)

`-47`

(C)

`49`

(D)

`-49`

Solution:

By using the condition that, if two circles are
intersect orthogonally, then

`2(g_1g_2 + f_1f_2) = c_1 + c_2`

Where `g_1 = -1, f_1 = 11, c_1 = 5`

and `g_2 = 7, f_2 = 3, c_2 = k`

=> `2(-1xx7 + 11xx3) = 5 + k`

=> `2(26) = 5 + k`

=> `k = 47`
Correct Answer is `=>` (A) `47`
Q 1580734617

The radius of the circle

`x^2 + y^2 + 4x + 6y + 13 = 0` is :
BITSAT 2006
(A)

`sqrt26`

(B)

`sqrt13`

(C)

`sqrt23`

(D)

`0`

Solution:

Given equation is

`x^2 + y^2 + 4x + 6y + 13 = 0`

or `(x^2 + 4x + 4) + (y^2 + 6y + 9) + 13 = 4 +9`

or` (x + 2)^2+ (y + 3)^2 = 0`

or `(x +2)^2 +(y+3)^2 = 0`

`therefore` Radius of circle = 0
Correct Answer is `=>` (D) `0`
Q 1510834710

The centre of the circle `x = 2 + 3 cos theta,
y = 3 sin theta- 1` is :
BITSAT 2006
(A)

`(3 ,3)`

(B)

`(2, -1)`

(C)

`(-2, 1)`

(D)

`(-1, 2)`

Solution:

Given parametric equations are

`x =2 + 3costheta, y = 3sintheta -1`

or `costheta = (x-2)/3, sintheta = (y +1)/3`

Since, `sin^2theta + cos^2theta = 1`

=> `((x - 2)/3)^2 + ((y + 1)/3)^2 = 1`

=> `(x - 2)^2 + (y + 1)^2 = 3^2`

`therefore` Centre of circle is `(2, -1)`
Correct Answer is `=>` (B) `(2, -1)`
Q 1531612522

The condition foe a line `y = 2x + c` to touch the circle `x^2 + y^2 = 16` is :
BITSAT 2006
(A)

`c = 10`

(B)

`c^2 = 80`

(C)

`c = 12`

(D)

`c^2 = 64`

Solution:

if `y = mx + c` touches the circle

`x^2 + y^2 = a^2,` then `c^2 = a^2(1 + m^2)`

Now, the line `y = 2x + c ` touches the circle

`x^2 + y^2 = 16`, if

`therefore c^2 = 16(1 + 4) = 16 xx 5`

or `c^2 = 80`
Correct Answer is `=>` (B) `c^2 = 80`
Q 1531334222

If `x/alpha + y/beta = 1` touches the circle `x^2 + y^2 = a^2`

the point `(1/alpha, 1/beta)` lies on a/an
BITSAT 2005
(A)

straight line

(B)

Circle

(C)

parabola

(D)

ellipse

Solution:

Since the line `x/alpha + y/beta = 1` touches the circle `x^2 + y^2 = a^2.`

`:.` The perpendicular distance from centre · `(0, 0)` to the tangent = radius of the circle.

`=> (|-1|)/sqrt(1/alpha^2 + 1/beta^2) = a`

`=> 1/a^2 =1/alpha^2 + 1/beta^2`

The locus of `(1/alpha, 1/beta)` is

`1/a^2= 1/x^2+ 1/y^2`

`:.` It represent a circle
Correct Answer is `=>` (B) Circle
Q 1501834728

Let `P (x_1, y_1 )` and `Q (x_2, y_2)` are two points such that their abscissa `x_1` and `x_2` are the roots of the equation `x^2 + 2x- 3 = 0` while the ordinate `y_1` and `y_2` are the roots of the equation `y^2 + 4y - 12 = 0`.

The centre of the circle with `PQ` as diameter is :
BITSAT 2005
(A)

`(- 1, - 2)`

(B)

`(1,2)`

(C)

`(1, - 2)`

(D)

`(- 1,2)`

Solution:

Given `x_1, x_2` are the roots of the equation

`x^2 + 2x - 3 = 0 `

`=> x^2 + 3x - x - 3 = 0`

`=> x (x + 3) - 1 (x + 3) = 0`

` => (x - 1) (x+ 3)=0`

` => x_1= - 3, x_2 = 1`

and `y_1, y_2` are the roots of the equation

`y^2 + 4y - 12 = 0`

`=> y ^2 + 6y - 2 y - 12 = 0`

`=> y (y + 6)-2 (y + 6) = 0`

`=> (y- 2) (y + 6) = 0`

`=> y_1 =- 6, y_2 =2`

`:.` Points are `P ( - 3, - 6)` and `Q (1, 2)`.

Since, `P` and `Q` are the end points of a diameter.

`:.` Centre = mid point of `PQ`

`= ((-3+1)/2, (-6 +2 )/2)`

`= (-1, -2)`
Correct Answer is `=>` (A) `(- 1, - 2)`
Q 1672423336

For a circle `x^2+y^2 =81 ` what is the equation of chord whose mid point is` ( -2, 3)`
UPSEE 2016
(A)

`2x-3y-13 =0`

(B)

`2x+3y +13 =0`

(C)

`2x-3y+13 =0`

(D)

`3x-2y+13 =0`

Solution:

Equation of chord ` T=S'`

`T =-2x+3y-81`

`S' =4+9-81 =-68`

`-2x+3y-81 -68`

`2x-3y+13 =0`
Correct Answer is `=>` (C) `2x-3y+13 =0`
Q 1682823737

The equation of that diameter of the
circle `x^2+y^2-6x+2y-8 =0` which
passes through the origin is

UPSEE 2016
(A)

`6x-y =0`

(B)

`3x+2y =0`

(C)

`x+3y =0`

(D)

`3x-y =0`

Solution:

The centre `C` of the circle is given by `[-1/2(-6),-1/2(2)]` or `(3,-1)`

Required diameter is the line joining the origin `(0,0)`and the centre `C(3,-1)` and hence the requires equation is

`y-0 =(-1-0)/(3-0)(x-0)`

`3y =-x`

`=> x+3y =0`
Q 2471401326

If `OA` and `OB ` are the tangents to the circle
`x^2 + y^ 2 -6x -8y+ 21 =0` drawn from the
origin `O`, then `AB` is equal to
UPSEE 2015
(A)

`11`

(B)

`4/5 sqrt 21`

(C)

`sqrt(17/3)`

(D)

None of these

Solution:

Given equation of circle is

`x^ 2 + y^ 2- 6x- 8y+ 21 = 0,`

whose centre =` (3, 4)`

and radius `= sqrt(9 +16- 21) = 2`

Clearly, the line `AB` is chord of contact and its
equation is

`x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = 0`

Here, `(x_1,y_1)=(0,0)`

`:. 0 + 0-3 (x + 0)- 4 (y + 0) + 21 = 0`

`=> 3x+4y-21=0` ............(i)

Now, perpendicular distance from `(3, 4)` to the line
(i) is

`CM=(3(3)+4(4)-21)/(sqrt(9+16))=4/5`

`AM=sqrt(AC^2-CM^2)=sqrt(4-16/25)=2/5sqrt 21`

`:. AB=2AM=4/5 sqrt 21`
Correct Answer is `=>` (B) `4/5 sqrt 21`
Q 2401734628

If the line `x + 2ky + 3 = 0` is a diameter of the
circle `x^2 + y^ 2- 6x + 2y = 0,` then `k` is equal
to
UPSEE 2015
(A)

`3`

(B)

`-5`

(C)

`-1`

(D)

`5`

Solution:

Given equation of circle is

`x^2+y^2-6x+2y=0`, whose

centre `= (3,-1)` and radius `= sqrt(9+1)=sqrt 10`

Clearly, the centre lies on the line

`x+2ky+3=0`

We have, `3-2k+3=0`

`x+2ky+3=0`

We have `3-2k+3=0`

`=> 6-2k=0`

`:. k=3`
Correct Answer is `=>` (A) `3`
Q 2415591469

The tangent at `(1, 7)` to the curve `x^2 = y - 6`
touches the circle `x^2 + y^2 + 16x + 12y + c = 0`
at
UPSEE 2014
(A)

`(6, 7)`

(B)

`(- 6, 7)`

(C)

`(6, -7)`

(D)

`(- 6, -7)`

Solution:

The tangent at `(1 , 7)` to the parabola `x^2 = y - 6` is

`x = 1/2 (y + 7) - 6`

`=> 2x = y + 7 - 12`

`=> y = 2x + 5`

which is also a tangent to the circle

`x^2 + y^2 + 16x + 12 y + c = 0`

`:. x^2 + (2x + 5)^2 + 16x + 12 (2x + 5) + c = 0`

`=> 5x^2 + 60x + 85 + c = 0`

must have equal roots.

Let `alpha` and `beta` be the roots of the equation.

Then, `alpha + beta = -12 => alpha = -6 quad ( ∵ alpha = beta)`

`:. x = - 6` and `y = 2x + 5 = - 7`

`=>` Point of contact is `(- 6, -7)`.
Correct Answer is `=>` (D) `(- 6, -7)`
Q 2416012879

The parametric equations of the circle
`x^2 + y^2 + mx + my = 0` are
UPSEE 2014
(A)

` x = - m/2 + m/sqrt2 cos theta , y = m/2 + m/sqrt(2) sin theta`

(B)

` x = - m/2 + m/sqrt2 cos theta , y = - m/2 + m/sqrt(2) sin theta`

(C)

` x = 0, y = 0`

(D)

None of the above

Solution:

Here, `x^2 + y^2 + mx + my = 0`

`=> (x^2 + mx ) + (y^2 + my )= 0`

` => (x^2 + mx + m^2/4 )^2 + (y^2 + my + m^2/4 )= m^2/2 `

` => [ x - ( - m/2)]^2 + [ y - ( - m/2)]^2 = ( m/sqrt(2))^2`

So, the parametric equations of the circle are

`x =- m/2 + m/sqrt(2) cos theta`

and ` y =- m/2 + m/sqrt(2) sin theta`
Correct Answer is `=>` (B) ` x = - m/2 + m/sqrt2 cos theta , y = - m/2 + m/sqrt(2) sin theta`
Q 2523178041

If `A= {(x,y) : x^2 +y^2 le1 ; x , y ϵ R }` and `B = {(x, y) : x^2 + y^2 le 4; x, y ϵ R}`, then
UPSEE 2013
(A)

`A -B = phi`

(B)

`B -A = phi`

(C)

`A nn B ne phi`

(D)

`A nn B = phi`

Solution:

A is the set of all points on the inner circle `x^2+y^2 = 1` `B` is the set of all points on the outer circle `x^2+y^2 = 4`




`therefore A - B = A , B -A = B , A nn B = phi`
Correct Answer is `=>` (D) `A nn B = phi`
 
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