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# Test: 35 Year JEE Previous Year Questions: Circle

## 33 Questions MCQ Test Additional Documents and Tests for JEE | Test: 35 Year JEE Previous Year Questions: Circle

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QUESTION: 1

### A triangle with vertices (4, 0), (–1, –1), (3, 5) is

Solution:

In isosceles triangle side AB = CA For right angled triangle, BC2 = AB2 + AC2

So, here BC = or BC2 = 52

or

QUESTION: 2

### Locus of mid point of the portion between the axes of x cosα + y sina = p whre p is constant is [2002]

Solution:

Equation of AB is x cosα + y sinα = p;

So co-ordinates of A and B are

So coordinates of  midpoint of AB are

⇒ cosα = p/2x1 and sinα = p/2y1  ;

cos2α + sin2α =  1

Locus of (x1, y1) is

QUESTION: 3

### If the pair of lines ax2+2hxy+by2+2gx+2fy+c=0 intersect on the y-axis then [2002]

Solution:

Put x = 0 in the given equation ⇒ by2 + 2 fy + c = 0.
For unique point of intersection f2– bc = 0 ⇒ af2 – abc = 0.
Since  abc + 2fgh – af2 –bg2 – ch2 = 0 ⇒ 2fgh – bg2 – ch2 = 0

QUESTION: 4

The pair of lines represented by 3ax2 + 5xy + (a2 – 2)y2 = 0 are perpendicular to each other for [2002]

Solution:

3a + a2 – 2 = 0 ⇒ a2 + 3a – 2 = 0.;

QUESTION: 5

A square of side a lies above the  x-axis  and has one vertex  at the origin. The side passing through the origin makes an angle   with the positive direction of x-axis.  The equation of its diagonal not passing through the origin is

Solution:

Co-ordinates of  A = (a cosα , a sinα) Equation of OB,

∴ slope of CA =

Equation of CA

y - a sinα =  (x - a cosα)

⇒ ( y - a sinα) (1 + tan a) = (a cosα - x )(1 - tana)
⇒ (y - a sinα) (cosα+ sinα) = (a cosα- x )(cosα- sinα)
⇒ y(cos+ sinα) - a sinα cosα - a sin2α = a cos2α - a cosα sinα- x (cosα- sinα)
⇒ y(cosα + sinα) + x(cosα - sinα)= a
y(sinα + cosα ) + x(cosα - sinα ) = a.

QUESTION: 6

If the pair of straight lines x2 - 2 pxy - y2 = 0 and x2 - 2qxy -y2=0 be such that each pair bisects the angle between the other pair, then [2003]

Solution:

Equation of bisectors of second pair of straight lines

is,    qx2 + 2 xy - qy2=0 ........(1)

It must be identical to the first pair
x2 - 2 pxy -y2=0 ........(2)

from (1) and  (2)

QUESTION: 7

Locus of centroid of the triangle whose vertices are (a cos t, a sin t ), (b sin t, - b cos t ) and (1, 0), where t is a parameter, is [2003]

Solution:

QUESTION: 8

If x1, x2 , x3 and y1, y2,y3 are both in G.P. with the same common ratio, then the points ( x1, y1), (x2 , y2) and (x3 , y3) [2003]

Solution:

Taking co-ordinates as

Then slope of line joining

and slope of line joining (x, y) and (xr, yr)

∴ m1 =  m2

⇒ Points lie on the straight line.

QUESTION: 9

If the equation of the locus of a point equidistant from  the point (a1, b1) and (a2, b2) is (a1 -b2) x + (a1 -b2) y + c= 0 ,  then the value of `c` is                             [2003]

Solution:

QUESTION: 10

Let A(2, - 3) and B ( -2, 3) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y= 1, then the locus of the vertex C is the line

Solution:

Let the vertex C be (h, k), then the

centroid of Δ ABC is

orIt lies on 2x + 3y  = 1

.

= Locus of  C is 2x + 3y = 9

QUESTION: 11

The equation of the straight line passing th rough the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is [2004]

Solution:

Let the required line be ......(1)

then a + b = –1 .......(2)

(1) passes through (4, 3) ,

⇒ 4b +3a= ab .................(3)

Eliminating b from (2) and (3), we get a2 -4 = 0

⇒ a =±2 ⇒b =-3 or1

∴ Equations of straight lines are

QUESTION: 12

If the sum of t he slopes of the lines given by x2 - 2cxy - 7y= 0 is four times their product c has the value[2004]

Solution:

Let the lines be y = m1x and y = m2x then

m1 + m2 = and  4 m1m2 =

Given m1 + m2= 4 m1m2

QUESTION: 13

If one of the lines given by 6x2 - xy + 4cy= 0 is 3x + 4y = 0, then c equals [2004]

Solution:

3x + 4y=0 is one of the lines of the pair

6 x2 - xy + 4cy2=0 , Put

we get

QUESTION: 14

The line parallel to the x- axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0, where (a, b) ≠ (0, 0) is [2005]

Solution:

The line passing through the intersection of lines
ax + 2by = 3b=0 and bx - 2ay - 3a=0
is ax + 2by + 3b + λ (bx – 2ay – 3a) = 0
⇒ (a + b λ) x + (2b – 2aλ)y + 3b – 3λa = 0
As this line is parallel to x-axis.
∴ a + bλ = 0  ⇒ λ = – a/b

⇒ ax + 2by + 3b – (bx – 2ay – 3a) = 0

⇒ ax + 2by + 3b – ax

So it is 3/2 units below x-axis.

QUESTION: 15

If a vertex of a triangle is (1, 1) and the mid points of two sides through this vertex are (–1, 2) and (3, 2) then the centroid of the triangle is [2005]

Solution:

Vertex of tringle is (1, 1) and midpoint of sides through this vertex is (– 1, 2) and (3, 2)

⇒ vertex B and C come out to be (– 3, 3) and (5, 3)

∴ Centroid is

QUESTION: 16

A straight line through the point A (3, 4) is such that its intercept between the axes is bisected at A. Its equation is

Solution:

∵ A is the mid point of PQ , therefore

a = 6, b = 8

∴ Equation of line is or 4x + 3y = 24

QUESTION: 17

If (a,a2) falls inside the angle made by the lines y =  x > 0 and y = 3x , x > 0 , then a belong to [2006]

Solution:

Clearly for point P,

a2 - 3a<0 and

QUESTION: 18

Let A (h, k), B(1, 1) and C (2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1square unit, then the set of values which 'k' can take is given by [2007]

Solution:

Given : The vertices of a right angled triangle A(l, k), B(1, 1) and C(2, 1) and Area of ΔABC = 1 square unit

We know that, area of right angled triangle

⇒ ± (k- 1)=2 ⇒ k = – 1, 3

QUESTION: 19

Let P = (–1, 0), Q = (0, 0) and R = (3, ) be three point. The equation of the bisector of the angle PQR is [2007]

Solution:

Given : The coordinates of points P, Q, R are (–1, 0),

(0, 0), respectively..

Slope of QR  =

⇒

Let QM bisects the ∠PQR ,

∴Slope of the line QM =

∴ Equation of line QM is (y – 0) =  -

QUESTION: 20

If one of the lines of my2 + (1– m2) xy – mx= 0 is a bisector of the angle between the lines xy = 0, then m is [2007]

Solution:

Equation of bisectors of lines, xy = 0 are y = ± x

∴ Put y = ± x in the given equation my2 + (1 – m2)xy – mx2 = 0
∴ mx2 + (1 – m2)x2 – mx2 = 0
⇒ 1 – m2 = 0 ⇒ m = ± 1

QUESTION: 21

The perpendicular bisector of the line segment joining P (1, 4) and Q(k, 3) has y-intercept –4. Then a possible value of k is[2008]

Solution:

Slope of

∴ Slope of perpendicular bisector of  PQ = ( k –1)

Also mid point of

∴ Equation of perpendicular bisector is

⇒ 2y – 7 = 2(k –1) x –(k2 –1)

⇒ 2(k – 1)x – 2y + ( 8 – k2) = 0

∴ y-intercept

⇒ 8 – k2 = –8  or k2 = 16  ⇒ k = ± 4

QUESTION: 22

The shortest distance between the line y – x = 1 and the curve x = y2 is : [2009]

Solution:

Let (a2, a) be the point of shortest distance on x = y2

Then distance between (a2, a) and line x – y + 1 = 0 is given by

It is min when

QUESTION: 23

The lines p(p2 +1)x – y + q = 0 and (p2 + 1)2x + (p2 + 1)y + 2q = 0 are perpendicular to a common line for : [2009]

Solution:

If the lines p (p2 + 1) x – y + q = 0
and (p2 + 1)2 x + (p2 + 1) y +2q = 0
are perpendicular to a common line then these lines must be parallel to each other,

∴ m1 = m2

⇒ (p2 + 1)  ( p + 1) = 0
⇒ p = – 1
∴ p  can have exactly one value.

QUESTION: 24

Three distinct points A, B and C are given in the 2-dimensional coordinates plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point (–1, 0) is equal to  . Then the circumcentre of the triangle ABC is at the point: [2009]

Solution:

Given that

P (1, 0), Q (– 1, 0)  and

⇒ 3AP = AQ
Let A = (x, y) then 3AP = AQ ⇒ 9 AP2= AQ2
⇒ 9 (x – 1)2 + 9y2 = (x + 1)2 + y2
⇒ 9 x2 – 18x + 9 + 9y2 = x2 +2x +1 + y2
⇒ 8x2 – 20x + 8y2 + 8 = 0

....(1)

∴ A lies on the circle given by eq (1). As B and C also follow the same condition, they must lie on the same circle.

∴ Centre of circumcircle of Δ ABC

QUESTION: 25

The line L given by   passes through the point (13, 32). The line K is parallel to L and has the equation      Then the distance between L and K is [2010]

Solution:

Slope of line L =

Slope of line K =

Line L is parallel to line k.

(13, 32) is a point on L.

Equation of K : y - 4x=3 ⇒ 4x -y + 3=0

Distance between L and K  =

QUESTION: 26

The lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line L3 : y + 2 = 0 at P and Q respectively. The bisector of the acute angle between L1 and L2 intersects L3 at R.
Statement-1: The ratio PR : RQ equals
Statement-2: In any triangle, bisector of an angle divides the triangle into two similar triangles.     [2011]

Solution:

L1 : y – x = 0
L2 : 2x + y = 0
L3 : y + 2 = 0
On solving the equation of line L1 and L2 we get their point of intersection (0, 0) i.e., origin O.
On solving the equation of line L1 and L3, we get  P = (– 2, – 2).
Similarly, we get Q = (– 1, – 2) We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [Angle Bisector Theorem of a Triangle]

QUESTION: 27

If the line 2x + y = k passes through the point which divides the line segment joining the points (1,1) and (2,4) in the ratio 3 :2, then k equals :     [2012]

Solution:

Let the joining points be A (1,1) and B (2,4).
Let point C divides line AB in the ratio 3 : 2.
So, by section formula we have

Since Line 2x + y = k passes through

∴ C satisfies the equation 2x + y = k.

QUESTION: 28

A ray of light along gets reflected upon reaching x-axis, the equation of the reflected ray is [JEE M 2013]

Solution:

Suppose B(0, 1) be an y point on given line and co-ordinate of A is   So, equation of

Reflected Ray is

QUESTION: 29

The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is : [JEE M 2013]

Solution:

From the figure, we have

a = 2, b = , c = 2
x1 = 0, x2 = 0, x3 = 2

Now, x-co-ordinate of incentre is given as

⇒ x-coordinate of incentre =

QUESTION: 30

Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). The equation of the line passing through (1, –1) and parallel to PS is:         [JEE M 2014]

Solution:

Let P, Q, R, be the vertices of ΔPQR

Since PS is the median, S is mid-point of QR

So,

Now, slope of PS

Since, required line is parallel to PS therefore slope of required line = slope of PS Now, eqn of line passing through (1, –1)  and having slope  is

y-(-1) =

9y + 9 = –2x + 2 ⇒ 2x + 9y + 7 = 0

QUESTION: 31

Let a, b, c an d d be n on -zer o numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d =0 lies in the fourth quadrant and is equidistant from the two axes then         [JEE M 2014]

Solution:

Given lines are

4ax + 2ay + c = 0
5bx + 2by + d = 0

The point of intersection will be

∵ Point of intersection is in fourth quadrant so x is positive and y is negative.
Also distance from axes is same So x = – y (∵ distance from x-axis is –y as y is negative)

QUESTION: 32

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is :         [JEE M 2015]

Solution:

Total number of integral points inside the square OABC = 40 × 40 = 1600
No. of integral points on AC

= No. of integral points on OB

= 40 [namely (1, 1), (2, 2) ... (40, 40)]

∴ No. of integral points inside the ΔOAC

QUESTION: 33

Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (–1, –2), then which one of the following is a vertex of this rhombus? [JEE M 2016]

Solution:

Let other two sides of rhombus are
x – y + λ = 0 and
7x –y + μ = 0
then O is equidistant from AB and DC and from AD and BC

and

∴ Other two sides are x – y – 3 = 0 and 7x – y + 15 = 0 On solving the eqns of sides pairwise, we get

the vertices as