Practice question on Circle for IIT-JEE , NDA and airforce

Circle

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Question no. 1

The area of the circle whose center is at (1,2) and passes through the point (4,6) is

looks_one 5π

looks_two 10π

looks_3 25π

looks_4 none of these

Option looks_3 25π

Solution :
center (1,2)

Question no. 2

If a circle passes through the point (0,0) , (a,0),(0,b) , then its center is

looks_one (a, b)

looks_two (b, a)

looks_3 (a/2, b/2)

looks_4 (b/2, -a/2)

Option looks_3 (a/2, b/2)

Solution :
General Equation of circle : x² + y² + 2gx + 2fy + c = 0
At (0 ,0)
0² + 0² + 2(0)x + 2(0)y + c = 0 ⇒ c = 0
At (a, 0)
a² + 2ga + 0 = 0 ⇒ g = -a/2
At (0, b)
b² + 2gb = 0
f = -b/2
center - (-g , -f)
put the value of g and f
center = (a/2 , b/2)

Question no. 3

The Circle x² + y² +4x -4y +4 = 0 touches

looks_one x-axis

looks_two y-axis

looks_3 x-axis and y-axis

looks_4 none of these

Option looks_3 x-axis and y-axis

Solution :
x² + y² +4x -4y +4 = 0
x² +4x + 4 - 4 + y² -4y +4 = 0
(x+2)² -4 + (x-2)² = 0
(x+2)² + (x-2)² = 4
center = (-2 , 2) and radius = 2 , it will touch both axes

Question no. 4

A circle touches the y-axis at the point (0,4) and cuts the x-axis in a chord of length 6 units. The radius of the circle is

looks_one 3

looks_two 4

looks_3 5

looks_4 6

Option looks_3 5

Solution :

Question no. 5

If the lines 3x-4y+4=0 and 6x-8y-7=0 are tangents to a circle, then the radius of the circle is

looks_one 3/2

looks_two 3/4

looks_3 1/10

looks_4 1/20

Option looks_two 3/4

solution :
the lines 3x-4y+4=0 and 6x-8y-7=0 are parallel to each other it means distance between them is diameter

Question no. 6

If the radius of the circle x² + y² -18x +12y +k = 0 be 11 , then k=

looks_one 347

looks_two 4

looks_3 -4

looks_4 49

Option looks_3 -4

Solution :

Question no. 7

Center of circle (x-x₁)(x-x₂) +(y-y₁)(y-y₂)=0 is

looks_one (x₁ + y₁)/2 , (x₂ + y₂)/2

looks_two (x₁ - y₁)/2 , (x₂ - y₂)/2

looks_3 (x₁ + x₂)/2 , (y₁ + y₂)/2

looks_4 (x₁ - x₂)/2 , (y₁ - y₂)/2

Option looks_3 (x₁ + x₂)/2 , (y₁ + y₂)/2

Solution :
(x-x₁)(x-x₂) +(y-y₁)(y-y₂)=0
(x₁,y₁) , (x₂ , y₂) are the end points of the diameter of the circle
center will be (x₁ + x₂)/2 , (y₁ + y₂)/2

Question no. 8

The number of circles touching the line y-x=0 and the y-axis is

looks_one zero

looks_two one

looks_3 two

looks_4 infinite

Option looks_4 infinite

Solution :
The number of circles touching the line y-x=0 and the y-axis is infinite

Question no. 9

If one end of a diameter of the circle x² +y² -4x -6y +11 = 0 be (3,4), then the other end is

looks_one (0,0)

looks_two (1,1)

looks_3 (1,2)

looks_4 (2,1)

Option looks_3 (1,2)

solution :
x² +y² -4x -6y +11 = 0
(x-2)² + (y-3)² =2
center = 2, 3
diameter one point = (3,4)
diameter other point = (x,y)
(3 +x)/ 2 = 2 ⇒ x = 1
(4 + y)/2 = 3 ⇒ y = 2

Question no. 10

Circle x² +y² +6y = 0 touches

looks_one y-axis at the origin

looks_two x-axis at the origin

looks_3 x-axis at the point (3,0)

looks_4 The line y+3=0

Option looks_two x-axis at the origin

Solution :
x² +y² +6y = 0
x² + (y +3)² = 9
center = (0 ,-3)

Question no. 11

For the circle x² +y² +6x -8y +9 = 0 , which of the following statements is true

looks_one circle passes through the point (-3,4)

looks_two circle through x-axis

looks_3 circle through y-axis

looks_4 none of these

Option looks_two circle through x-axis

Solution :
x² +y² +6x -8y +9 = 0
(x+3)² + (y-4)² = 4²
center - (-3,4)
circle through x-axis

Question no. 12

The equation of the circle which touches the lines x=0, y=0 and 3x+4y=4 is

looks_one x² +y² - 4x + 4y +4 =0

looks_two x² +y² + 4x + 4y +4 =0

looks_3 x² +y² - 4x - 4y +4 =0

looks_4 x² +y² + 4x - 4y +4 =0

Option looks_3 x² +y² - 4x - 4y +4 =0

Solution :
Given line is the passing through the first quadrant , therefore center must be in first quadrant

Question no. 13

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

looks_one x² +y² +4x +4y -5 =0

looks_two x² +y² -4x -4y -5 =0

looks_3 x² +y² -4x=13

looks_4 x² +y² -4x -4y +5 =0

Option looks_two x² +y² -4x -4y -5 =0

Solution :
r² = (4-2)² + (5-2)² = 2² + 3² = 13
equation of circle - (x-x₁)² + (y - y₁)² = r²
(x-2)² + (y-2)² = 13 ⇒ x² +y² -4x -4y -5 =0

Question no. 14

Radius of circle (x-5)(x-1) + (y-7)(y-4)=0 is

looks_one 3

looks_two 4

looks_3 5/2

looks_4 7/2

Option looks_3 5/2

Solution :
(x-5)(x-1) + (y-7)(y-4) = 0 is the type of equation (x-x₁)(x- x₂) + (y -y₁)(y-y₂) = 0 where (x₁ , y₁ ) (x₂ ,y₂) are the points of the diameter points
(5,7) and (1,4) are the end point of diameter points
(2r)² = (5-1)² + (7-4)² = 25
2r = 5
r = 5/2

Question no. 15

The equation of the circle with centre at (1,-2) and passing through the center of the given circle x² + y² +2y -3=0 , is

looks_one x² +y² -2x +4y +3 =0

looks_two x² +y² -2x +4y -3 =0

looks_3 x² +y² +2x -4y -3 =0

looks_4 x² +y² +2x -4y +3 =0

Option looks_one x² +y² -2x +4y +3 =0

Solution:
center - (1,-2)
equation of circle - (x-1)² + (y+2)² = r²
Center of the circle x² + y² +2y -3=0 is (0, -1)
this point will satisfy the eqaution of above circle (x-1)² + (y+2)² = r²
(0-1)² + (-1+2)² = r²
r² = 2
Equation of circle - (x-1)² + (y+2)² = 2
x² +y² -2x +4y +3 =0

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