Free Objective Test 02 Practice Test - 11th and 12th 

Question 1

If (mi,1mi), i = 1, 2, 3, 4 are con - cyclic points, then the value of m1m2m3m4 is 

A.

1

B.

-1

C.

0

D.

2

SOLUTION

Solution : A

Let equation of circle be x2+y2 + 2gx + 2fy + c = 0. If (m,1m) lies on this circle, then

m2+1m2+2gm+2f1m+c=0

or m4+2gm3+2fm+cm2+1=0

This is a fourth degree equation in m having m1,m2,m3,m4 as its roots.

Therefore, m1m2m3m4 = product of roots = 11 = 1.

Question 2

The equation of the circle whose centre is (1, -3) and which touches the line 2x - y - 4 = 0 is

A. 5x2+5y210x+30y+49=0
B. 5x2+5y2+10x30y+49=0
C. 5x2+5y210x+30y49=0
D. None of these

SOLUTION

Solution : A

Radius of circle the required circle =2+345=15
Therefore the equation is,
(x1)2+(y+3)2=15
i.e, x2+y22x+6y+1+9=15
i.e, 5x2+5y210x+30y+49=0.

Question 3

The centres of the circles x2+y2=1,  x2+y2+6x2y=1 and x2+y212x+4y=1 are

A. same
B. Collinear
C. Non–collinear
D. None of these

SOLUTION

Solution : B

Centre are (0, 0), (-3, 1) and (6, -2) and a line passing through any two points say (0, 0) and (-3, 1) is y=13x and point (6, -2) lies on it. Hence points are collinear.

Question 4

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

A. 5π
B. 10π
C. 25π
D. None of these

SOLUTION

Solution : C

Radius = (14)2+(26)2=5
Hence the area is given by πr2=25π sq. units.

Question 5

If a circle whose centre is (1, –3) touches the line 3x – 4y –5 = 0, then the radius of the circle is

A. 2
B. 4
C. 52
D. 72

SOLUTION

Solution : A

Radius = perpendicular distance from (1, -3) to the given line 3x - 4y - 5 = 0.
i.e. 3+12552=2.

Question 6

The locus of the centre of a circle which touch the circles x2+y26x6y+14=0 externally and also the Y - axis is given by

A.

x26y7y+14=0

B.

x210x6y+14=0

C.

y26x10y+14=0

D.

y210x6y+14=0

SOLUTION

Solution : D

Let (x, y) be centre of circle which touch y - axis and given circle. 
At any point, the radius of this circle will be equal to 'x' units. Since this circle is touching the given circle which has a radius of two units,
Distance between centres of the two circles = 2 + x

(x3)2+(y3)2=2+x

y210x6y+14=0

Question 7

If a straight line through C(8,8) making an angle 135 with the x-axis cuts the circle x=5cosθ,y=5sinθ in points A and B, then length of segment AB is

A. 5
B. 10
C. 15
D. 152

SOLUTION

Solution : B

Inclination of the line AB is 135Slope=tan135=1
Equation of AB is y8=1(x+8)x+y=0
x + y = 0 passes through the centre of the circle x2+y2=25
Length of the chord AB = Diameter of the circle =2×5=10

Question 8

If P(2,8) is an interior point of a circle x2+y22x+4yp=0 which neither touches nor intersects the axes, then set for p is -

A. p<1
B. p<4
C. p>96
D. pϕ

SOLUTION

Solution : D

Since the point P is interior to the circle, S1 < 0
 =(22)+(82)2.(2)+4(8)p<0 
 =96p<0 
 =p>96
Also given that the circle doesn't touches any of the axes.
So, g2c < 0
      f2c < 0
g2c < 0 
 =1+p<0
 =p<1
Also, 
f2c < 0
 =4+p<0
 =p<4
Since  p<4 and p>96 is not possible, the set for p will be a null set meaning it'll not have any element in it.

Question 9

Tangents are drawn from any point on the circle x2+y2=R2 to the circle x2+y2=r2. If the line joining the points of intersection of these tangents with the first circle also touch the second, then R equals

A. 2r
B. 2r
C. 2r23
D. 4r35

SOLUTION

Solution : B


ΔABC is equilateral
Circum radius = 2 (in radius)
R = 2r               

Question 10

Two circles of radii 4 cms and 1 cm touch each other externally and θ is the angle contained by their direct common tangents. Then sin θ is =

A.

2425

B.

1225

C.

34

D.

none

SOLUTION

Solution : A

sinθ2=r1r2r1+r2=35,   cosθ2=45sin θ=2425

Question 11

The two circles x2+y2+2ax+c=0 and x2+y2+2by+c=0 touch if 1a2+1b2=

A.

1c2

B.

c

C.

1c

D.

c2

SOLUTION

Solution : C

We can write the equation of common tangent, with the help of radical axis.
Since two circles are touching each other the common tangent will be the radical axis, and the equation of the radical axis we can write - 
S = S'
x2+y2+2ax+c=x2+y2+2by+c=0
So, Equation of Common tangent is ax – by = 0.

Now, we can find the perpendicular distance from the center of the first circle to the tangent which also should be equal to radius of that circle. 
Perpendicular distance from (-a,0) to the tangent  = a2a2+b2
Radius of the same circle will be =  a2c
Therefore a2a2+b2=a2c

Question 12

Two circles, each of radius 5 units, touch each other at (1, 2). If the equation of their common tangent is 4x + 3y – 10 = 0 then equation of one such circle is

A.

x2+y26x+2y15=0

B.

x2+y210x10y+25=0

C.

x2+y2+6x2y15=0

D.

x2+y210x10y25=0

SOLUTION

Solution : B

 

Let the point of contact be (x1,y1) = (1, 2)
We know that co-ordinates of any point on the circle can be given by -
a±rcosθ, b±rsinθ)
Where (a,b) are the co-ordinates of the center of the circle and θ) is the angle made by the line joining that point and the center with the horizontal axis(X-axis).
We are the given the equation of the tangent.
We know that the radius is perpendicular to the tangent. So the slope of radius will be negative inverse of the slope of the tangent.
Slope of the tangent is 43
And slope of the radius will be -  34
This slope is nothing but the tan of the angle it makes with the horizontal.
 So, tan(θ) =  34
cos θ=45, sin θ=35
Now we can write -
(1,2 ) = a±5cosθ, b±5sinθ) 
(1,2 ) = a±4, b±3) 
On comparing we get - 
centre=(1±4,2±3)=(5,5)or(3,1)
Using this we can write the equation of circle.
x2+y210x10y+25=0
x2+y2+6x+2y15=0

Question 13

The equations to the circles which touch the lines 3x – 4y + 1 = 0, 4x + 3y – 7 = 0 and pass through (2, 3) are

A. x2 + y2 - 4x - 16y + 43 = 0, 5x2 + 5y2 - 12x - 24y + 31 = 0
B. x2 + y2 + 4x + 16y - 43 = 0, 5x2 + 5y2 - 12x - 24y + 31 = 0
C. x2 + y2 - 4x - 16y + 43 = 0, 5x2 + 5y2 + 12x + 24y + 31 = 0
D. x2 + y2 + 4x + 16y - 43 = 0, 5x2 + 5y2 + 12x + 24y + 31 = 0

SOLUTION

Solution : A

Since the circle touches the given lines which are perpendicular, centre of the circle lies on the bisectors of angels between the given lines.
The bisectors of angles between the given lines are 3x4y+15 ± 4x +3y75 = 0
(3x - 4y + 1) + (4x + 3y - 7) = 0 or (3x - 4y + 1) - (4x + 3y - 7) = 0
7x - y - 6 = 0 or - x - 7y + 8 = 0 y = 7x - 6 or x = 8 - 7y
If the centre lies on x = 8 - 7y, then the centre is of the form (8 - 7k, k)
3(87k)4k+15 = (87k2)2+(k3)2 (2425k+1)2=25(67k)2+25(k3)2
252(15)2=25[36+49k284k+k2+96k] 25(1+k22k)=50k290k+45
25k2 - 40k + 20 = 0 5k2 - 8k + 4 = 0 k is imaginary
The centre of the circle lies on the line y = 7x - 6 and hence it is of the form (k, 7k - 6)
3k  k(7k  6) + 15 = (k2)2+(7k63)2 (3k - 28k + 24 - 1)2 = 25 [(k - 2)2 + (7k - 9)2]
(25 - 25k)2 = 25(k2 + 4 - 4k + 49k2 + 81 - 126k) 25(1 + k2 - 2k) = 50k2 - 130k + 85
25k2 - 80k + 60 = 0 5k2 - 16k + 12 = 0 (k - 2) (5k - 6) = 0 k = 2 or k = 65
If k = 2, then centre = (2, 8), radius = 5
Equation of the circle is (x2)2+(y8)2 = 25 x2 + y2 - 4x - 16y + 43 = 0
If k = 65, then centre is (65,  125), radius = (45)2+(35)2 = 1
Equation of the circle is (x65)2 + (y125)2=1(5x6)2+(5y12)2 = 25
25x2 + 25y2 - 60x - 120y + 155 = 0 5x2 + 5y2 - 12x - 24y + 31 = 0
Required circles are x2 + y2 - 4x - 16y + 43 = 0 and 5x2 + 5y2 - 12x - 24y + 31 = 0

Question 14

The slope m of a tangent through the point (7, 1) to the circle x2+y2=25 satisfies the equation

A. 12m2 + 7m - 12 = 0
B. 12m2 + 7m + 9 = 0
C. 12m2 - 7m - 12 = 0
D. 9m2 + 12m + 16 = 0

SOLUTION

Solution : C

Equation of a tangent to x2+y2 = 25 is y = mx ± 51+m2
This passes through (7, 1) 1 = 7m ± 51+m2
(17m)2=25(1+m2) 49m214m+1=25+25m2 24m214m14=0
12m2 - 7m - 12 = 0

Question 15

The number of points with integral coordinates that are interior to the circle x2+y2=16 is

A. 43
B. 49
C. 45
D. 51

SOLUTION

Solution : C

We are asked to find the integral co-ordinates for (x, y) interior to the circle x2+y2=16  or satisfying x2+y2 < 16
Well, to answer this question we'll see all the integral coordiantes and check it by putting the values in the equation of the circle.
Co-ordinates will be - 
(0,0), (0,1),(0,2), (0,3)
 (1,0), (1,1),(1,2), (1,3)
 (2,0), (2,1),(2,2), (2,3)
 (3,0)  (3,1) (3,2) 
Out of these 15 points 4 points lie on the x- axis including origin and 3 points lie on the Y-axis excluding origin.
Therefore after excluding these points we have 8 points in the 1st quadrant which are inside the circle and not lying on any axis.
We can directly say the points inside all the four quadrants will be nothing but = 8*4 = 32
Now we have to account the points which are lying on axes.
On X- axis we have 7 points including origin.
On Y - axis we have 6 points excluding origin.
So total points will be =  32 + 7 + 6  = 45