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

=>

=>

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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)

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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.

hence

solving which we get k=

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

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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.

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.

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

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, )

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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.

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 (

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The equations x = a cos + b sin , and , 0 2 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+y2

we get a2+b2

=> x2+y2 =a2+b2

which is quadratic in nature,

coefficient of x2 = coefficient of y2

and no term involving xy

hence its locus represents a circle

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The line y = c is a tangent to the parabola x2 = 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

x2=c-1

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.

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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 =

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