Practice Question for Circle | NDA | IITJEE | AIRFORCE | SET 3 | 31-45 | UPSC |SAT

Circle

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Question no. 31

If the chord of contact of tangents drawn from a point on the circle x² + y² = a ² to the circle x² + y² = b² touches the circle x² + y² = c² , then a ,b ,c are in :

looks_one G.P.

looks_two H.P.

looks_3 A.P.

looks_4 none of these

looks_one G.P.

Solution :
The point (x₁ , y₁) from which the tangent are drawn is on circle x² + y² = a² , it means
x₁² + y₁² = a²
Equation of chord : xx₁ + y y₁ = b²
The chord will be the tangent to the circle x² + y² = c²
distance from the origin to that tangent is equal to radius 'c'
Distance of line (xx₁ + y y₁ = b²) from origin :

Question no. 32

what is the equation to circle which touches both the axes and has centre on the line x+y-4=0 ?

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

looks_3 x² + y² -4x -4y +4 = 0

Solution :
It touches both the axis , therefore
|g| = |f| = r
Center passes through the line x+y = 4
So the center will be in first quadrant
So (x-2)² + (y-2)² = 2² will be the equation.

Question no. 33

If the Circle x² + y² +2gx +2fy +c = 0 (c>0) touches the y-axis , then which one of the following is correct ?

looks_one g=-√c only

looks_two g= ∓√c

looks_3 f=√c only

looks_4 f=∓√c

looks_4 f=∓√c

Solution :

Circle touches the y-axis, therefore r = |g|
r²= g² + f² -c
(g)² = g² + f² -c
f² = c
f=∓√c

Question no. 34

The radius of the circle is x² + y² +2gx +2fy +c = 0 be r, then it will touch both the axes , if

looks_one g = f = r

looks_two g = f = c = r

looks_3 g=f = √c =r

looks_4 g=f and c²= r

Option looks_one g = f = r

Solution :
If circle touches both axes , then |g| = |f| = |r|

Question no. 35

The equation x² + y²=0 denotes

looks_one a point

looks_two a circle

looks_3 x -axis

looks_4 y- axis

Option looks_one a point

Solution :
It represent a point

Question no. 36

Equation of the circle passing through origin is x² + y² -6x +2y= 0 , what is the equation of one of its diameters?

looks_one x+3y=0

looks_twox+y=0

looks_3 x=y

looks_4 3x+y=0

Option looks_one x+3y=0

solution :
Equation of one of its diameter will pass through (0,0) and (3,-1)
y-0 = (0+1)x/(0-3)
y = -x/3
3y = -x
x+3y = 0

Question no. 37

Center of circle (x-3)² +(y-4)²=5 is

looks_one (3,4)

looks_two (-3,-4)

looks_3 (4,3)

looks_4 (-4,-3)

Option looks_one (3,4)

soution :
(x-g)² + (y-f)² = r²
So , the center is (g,f)
(x-3)² +(y-4)²=5 , the center is (3,4)

Question no. 38

The equation of the circle passing through the points (0,0) , (0,b) and (a ,b) is

looks_one x² + y² +ax +by= 0

looks_two x² + y² -ax +by= 0

looks_3 x² + y² -ax -by= 0

looks_4 x² + y² +ax -by= 0

Option looks_3 x² + y² -ax -by= 0

Solution :
equation = x² + y² + 2gx + 2fx + c = 0
At (0 , 0 )
x² + y² + 2gx + 2fx + c = 0
0 + 0 + 0 + 0 + c = 0
c = 0
At (0,b)
b² + 2fb=0
f = -b/2
At (a,b) x² + y² + 2gx + 2fx + c = 0
a² + b² + 2ga + 2(-b/2)b = 0
g = -a/2
put the value of g,f ,c in equation
equation of circle = x² + y² -ax -by= 0

Question no. 39

If the circle x² +y² +2gx +2fy +c = 0 is touched by y=x at P such that OP= 6√2, then the value of c is :

looks_one 36

looks_two 48

looks_3 72

looks_4 none of these

Option looks_3 72

Solution :

Apply Pythagoras theorem
OP² + (g² + f² - c) = distance of center from origin
(6√2)² + g² + f² - c = g² + f²
c = (6√2)² = 72

Question no. 40

The equation of circle passing through the points (4,3) and (3,2) and touching the line 3x-y-17=0

looks_one x² +y² + 31x - 43y -20 =0

looks_two x² +y² - 31x + 43y -20

looks_3 x² +y² + 9x + 3y +20

looks_4 x² +y² - 9x + 3y +20

option

very soon 10

Question no. 41

Find the equation of a circle touching the line x + 2y = 1 at the point (3,-1) and passing through the point (2,1).

looks_one x² +y² - 23x - 4y +35 =0

looks_two 3x² + 3y² - 4x - 23y +35 =0

looks_3 3x² + 3y² - 23x - 4y +35 =0

looks_4 None of these

option

very soon 11

Question no. 42

Find the equation of the circle which touches the Y-axis at a distance of +4 from the origin and cuts off an intercept of 6 units from the X- axis

looks_one x² +y² + 10x + 8y -16 =0

looks_two x² +y² + 10x - 8y -16 =0

looks_3 x² +y² - 10x - 8y +16 =0

looks_4 x² +y² - 10x + 8y +16 =0

option looks_3 x² +y² - 10x - 8y +16 =0

solution :

radius = 5
(x+g)² + (y+f)² = 25
(0 , 4) is point on circle and y coordinate of center is 4
it means f=-4
for g :
(x+g)² + (y+f)² = 25
(0 + g)² + (4-4)² = 25
g = 5
the equation is x² +y² - 10x - 8y +16 =0

Question no. 43

The angle between the tangents drawn from the origin to the circle (x-7)² + (y+1)² = 25 is

looks_one π/3

looks_two π/6

looks_3 π/2

looks_4 π/8

option looks_3 π/2

Solution :
radius = 5
length of tangents = √(S) = √((x-7)² + (y+1)² -25 ) = 5

Both the sides pf triangle are 5 , ∠A = 90^o , So ∠OAB = 45^o

Question no. 44

Find the angle between the circles : S₁ : x² + y² -4x + 6y +11=0 and S₂ : x² + y² -2x + 8y + 13=0

looks_one 60^o

looks_two 30^o

looks_3 45^o

looks_4 90^o

option looks_3 45^o

Solution :
S₁ : x² + y² -4x + 6y +11=0
radius(r₁) = √2 , center = (2 -3)
S₂ : x² + y² -2x + 8y + 13=0
radius(r₂) = 2 , center = (1,-4)
Distance between two center (AB) = √2
angle between the circle :

Question no. 45

Find the equation of the circle which passes through the origin and cuts orthogonally each of the circles : x² + y² - 8y +12=0 and x² + y² -4x - 6y -3=0

looks_one x² + y² + 6x +3y=0

looks_two x² + y² - 6x -3y=0

looks_3 x² + y² + 6x -3y=0

looks_4 x² + y² - 6x + 3y=0

option looks_3 x² + y² + 6x -3y=0

Solution :
If two circles cut orthogonally , then g₁g₂ + f₁f₂ = c₁ + c₂
Circle passes through origin therefore c= 0 g₁ = ? , f₁ = 0
Given data : x² + y² - 8y +12=0
g₂ = 0 , f₂ = -4
x² + y² -4x - 6y -3=0
g₃ = -2
f₃ = -3
For first pair of equation :
g₁g₂ + f₁f₂ = c₁ + c₂
2g₁*(0) + 2f₁(-4) = 0 + 12
f₁ = -3/2
For second pair of equation :
g₁g₂ + f₁f₂ = c₁ + c₂
2 g₁*(-2) + 2f₁*(-3) = 0 - 3
g₁ = 3
The equation will be x² + y² + 6x -3y=0

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