Topic: Conic Sections (Test 1)

Topic: Conic Sections
If the line 2x – y +  = 0 is a diameter of the circle  then  =
A. 9
B. 12
C. 3
D. 6
Answer : Option A
Explaination / Solution:

Equation of circle is 

Applying completing the square method

Comparing the above equation with   we get center as (-3,3) and radius as    .

As centre of the circlre lies on diameter , it will satisfy the equation of diameter, so on putting (-3,3) in equation of diameter we get



Circumcentre of the triangle, whose vertices are (0, 0), (6, 0) and (0, 4) is
A. ( 3, 2)
B. (2, 0)
C. (0, 3)
D. (3, 0)
Answer : Option A
Explaination / Solution:

circumcentre of a right angled triangle ABC right angled at A is  as circumcentre of right angled triangle lies on the mid pont of the hypotenuse.

so mid point of BC=(,) i.e.(3,2)

The value of k, such that the equation   represents a point circle, is
A. 252
B. 25252
C. 0
Answer : Option B
Explaination / Solution:

which gives radius  for a point circle radius should be zero.
solving which we get k=

The number of points on X-axis which are at a distance c units (c 3) from ( 2, 3) is
A. 3
B. 0
C. 2
D. 1
Answer : Option B
Explaination / Solution:

the shortest distance from x-axis to the point is 3.

The equation  represents a
A. pair of parallel straight lines
B. pair of perpendicular straight lines
C. pair of non-perpendicular intersecting straight lines
D. circle
Answer : Option C
Explaination / Solution:

On factorizing the given equation we get x-4y=0 and x-3y=0, which represents a pair of intersecting straight lines. Since the product of their slopes is not equal to -1, they are non perpendicular.


The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle  is

A. (12,2)
B. (12,32)
C. (12,12)
D. (12,+2)
Answer : Option A
Explaination / Solution:

Since the circle passes through (0,0) the equation reduces to 

c= 0 -----(1)

Since it passes through (1,0),

1 + 2g + c = 0

This implies g = -1/2

Since the circle touches the circle x2 + y2 = 9, their radii should be equal

2 = 3

Substituting the values and simplifying we get f = 

Hence the centre is (1/2, )

A circle with its centre on the line y = x + 1 is drawn to pass through the origin and touch the line y = x + 2. The centre of the circle is
A. (12,12)
B. (-1, 2)
C. (- 1, 0)
D. (12,12)
Answer : Option D
Explaination / Solution:

  be the equation of circle with (h,k) as the center and r be the radius.
As the center lies on the line y=x+1

=> k=h+1 or k=1+h --(i)

Circle passes through origin so (0,0) will satisfy the equatio of line so putting (0,0) in equation of circle we get 

putting k from (i) into (ii), we get

and as the circle touches the line y=x+2 so radius should be equal to the distance from the center to this line.

putting value ok k from (i) in above equation, we get

putting the value of r in (iii)

h= putting h in (i) we get k=

hence center is (

The equations x = a cos  + b sin  , and , 0   represent
A. a parabola
B. a hyperbola
C. an ellipse
D. a circle
Answer : Option D
Explaination / Solution:

x = a cos  + b sin  , and 

putting the value of x and y in x2+y

  we get   a2+b2

=> x2+y=a2+b2

which is quadratic in nature,

coefficient of x= coefficient of y2

and no term involving xy

hence its locus represents a circle

The line y = c is a tangent to the parabola x= y - 1 if c is equal to
A. 2 and 2
B. a
C. c = 1
D. 0
Answer : Option C
Explaination / Solution:

putting the value y=c into parabola,we get


or x2-(c-1)=0

here discriminat=

line y=c is tangent when discriminant is equal to 0.

putting disriminant =0 we get c=1.

(0, c) will be a point on the parabola. 

The eccentricity of the hyperbola  is
A. less than 1
B. None of these
C. √2
D. 1
Answer : Option C
Explaination / Solution:

above equation can be written as,

comparing it with the standard equation we get a=3 and b=3

as c=

we get c= 3 

and as e = 

we get e =