Free Objective Test 01 Practice Test - 11th and 12th
Question 1
The centre of the circle passing through (0,0) and (1,0) and touching the circle x2+y2=9 can be
(12,12)
(12,−√2)
(32,12)
(12,32)
SOLUTION
Solution : B
Let the centre of the circle be (x, y)
x=0+12
=12
r=32
√x2+y2=32
y=±√2
So there are two circles possible.
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Question 2
The length of common chord of the circles (x−a)2+y2 =a2 and x2+(y−b)2 = b2 is
2√a2+b2
ab√a2+b2
2ab√a2+b2
2ab√a2−b2
SOLUTION
Solution : C
Equation of common chord is ax−by = 0
Now length of common chord
= 2√r21−p21 = 2√r22−p22
Where r1 and r2 are radii of given circles and p1,p2 are the perpendicular distances from centres of circles to common chords.
Hence required length = 2√a2−a4a2+b2 = 2ab√a2+b2.
Question 3
The power of (2,1) with respect to the circle 2x2+2y2−8x−6y+k = 0 is positive if
0<k<12
-12<k<12
k>12
k<12
SOLUTION
Solution : C
2x2+2y2−8x−6y+k = 0
⇒x2+y2−4x−3y+(k2) = 0
4+1−8−3+k2>0
k2−6>0→k>12
Question 4
Equation of circle touching the line |x-2|+|y-3|=4 will be
(x−2)2+(y−3)2=12
(x−2)2+(y−3)2=4
(x−2)2+(y−3)2=10
(x−2)2+(y−3)2=8
SOLUTION
Solution : D
Perpendicular distance from centre to tangent = radius
r=|2+3−9|√2=4√2=4√22=2√2 Equation of circle is (x−2)2+(y−3)2=8
Question 5
If the line x + 2by + 7 = 0 is diameter of the circle x2+y2−6x+2y=0, then b =
SOLUTION
Solution : D
Here the centre of circle (3, -1) must lie on the line x + 2by + 7 = 0
Therefore, 3 - 2b + 7 = 0 ⇒ b = 5
Question 6
A circle with centre at the origin and radius equal to a meets the axis of x and A and B. P(α) and Q(β) are two points on this circle so that α−β=2γ, where γ is a constant. The locus of the point of intersection of AP and BQ is
x2−y2−2aytanγ=a2
x2+y2−2aytanγ=a2
x2+y2+2aytanγ=a2
x2−y2+2aytanγ=a2
SOLUTION
Solution : B
Coordinates of A are (–a, 0) and of P are (acosα,asinα)
∴ Equation of AP is y=asinαa(cosα+1)(x+a)⇒y=tan(α2)(x+a)→(1)
Similarly equation of BQ = y=asinβa(cosβ−1)(x−a)⇒y=−cot(β2)(x−a)→(2)
From (1) and (2), tan(α2)=yα+x,tan(β2)=a−xy
Now α−β=2γ
⇒tanγ=tan(α/2)−tan(β/2)(a+x)y+(a−x)y=yα+x−a−xy1+yα+x.a−xy⇒tanγ=tan(α/2)−tan(β/2)1+tan(α/2)tan(β/2)=x2+y2−a22ay⇒x2+y2−2aytanγ=a2 which is required locus.
Question 7
Two vertices of an equilateral triangle are (–1, 0) and (1, 0) and its third vertex lies above the x-axis, the equation of the circumcircle is
SOLUTION
Solution : A
Let A(–1, 0), B(1, 0) and C(0, b) be the vertices of the triangle, since C lies on the locus of points equidistance from A(–1, 0), and B(1, 0) i.e., y-axis.
Then AB=AC
⇒√1+b2=2
⇒b2=3
⇒b=√3 [∵ b>0]
Since the triangle is equilateral, the centre of the circumcircle is at the centroid of the triangle which is (0,1√3).
Thus the equation of the circumcenter is,
(x−0)2+(y−1√3)2=(1−0)2+(0−1√3)2
⇒x2+y2−(2√3)y+13=1+13
⇒3x2+3y2−2√3y=3
Question 8
The angle at which the circles (x−1)2+y2= 10 and x2+(y−2)2=5 intersect is
π6
π4
π3
π2
SOLUTION
Solution : B
r1=√10r2=√5d=√(1)2+(2)2=√5cosθ=r21+r22−d22r1r2=10+5−52√10√5=102×5×√2=1√2
Question 9
Consider a family of circles which are passing through the point (–1, 1) and are tangent to x-axis. (h, k) are the co-ordinates of the centre of the circles, then the set of values of k is given by the interval
SOLUTION
Solution : D
Since, circles are passing through the point (–1, 1) and touching X-axis, the equationof circles can be written as -
(h+1)2+(K–1)2=K2
h2+2h+2–2K=0
If h is real the Δ≥0
4- 4.1. (2 – 2K) ≥0
= 4 - 8 + 8k ≥0
= 8k ≥4
= k≥12
Question 10
The number of common tangents to the circles x2+y2+2x+8y−23=0 and x2+y2−4x−10y+19=0 is
1
2
3
4
SOLUTION
Solution : C
To find the no. of common tangents, let's find the relation between distance between the centers of these circles and sum of their radii -
Co-ordiantes of the center of first circle is (-1, - 4)
Co-ordiantes of the center of second circle is (2, 5)
thus the distance between them is - √(32+92)
= 3√(10)
Now, the sum their radii will be -
Radius of the first circle is - 2√(10)
Radius of the second circle is - √(10)
Sum of them will be - 3√(10)
Since, the distance between the centres is equal to sum of their radii
Number of common tangents will be 3.
Question 11
If the lengths of the tangents from two points A, B to a circle are 6, 7 respectively. If A, B are conjugate points then AB =
SOLUTION
Solution : B
Let the equation of the circle be x2 + y2=a2 and A (x1,y1), B(x2,y2)
A, B are conjugate points ⇒x1x2+y1y2=a2 → (1)
Length of the tangents from A, B are 6, 7 respectively ⇒ √x21+y21−a2 = 6, √x22+y22−a2= 7
⇒ x21+y21−a2 = 36
x22+y22−a2 = 49
⇒ x21+y21+x21+y22−2a2 = 85
⇒ x21+y21+x21+y22−2
(x1 x2 + y1 y2) = 85 [From (1)]
⇒ (x21+x22−2x1x2) + (y21+y22−2y1y2) = 85 ⇒ (x1−x2)2 + (y1−y2)2 = 85 ⇒ AB2 = 85 ⇒ AB = √85
Question 12
The equation of the circle passing through the point (1, –2) and having its centre on the line 2x – y – 14 = 0 and touching the line 4x + 3y – 23 = 0
SOLUTION
Solution : D
Let P(1, -2) and the centre be C(a, b)
Centre lies on 2x - y - 14 = 0 ⇒ 2a - b - 14 = 0 ⇒ b = 2a - 14 → (1)
Circle touches the lines 4x + 3y - 23 = 0. Lets say that this point of touching is Q.
We know that,
CQ = CP = Radius of the circle.
⇒ √(a−1)2+(b+2)2 = |4a + 3b − 23|√42 + 32
⇒ √(a−1)2+(2a−14+2)2 = |4a+3(2a−14)−23|5 [From (1)]
⇒ 5√a2−2a+1+4a2−48a+144 = |10a−65| ⇒ 5√5a2−50a+145 = 5|2a−13|
⇒ 5a2 - 50a + 145 = (2a - 13)2 = 4a2 - 52a + 169
⇒ a2 + 2a - 24 = 0 ⇒ (a + 6)(a - 4) ⇒ a = 4
Case 1 : a = 4 ⇒ b = 2(4) - 14 = - 6.∴ Centre C = (4, - 6)
Radius = CP = √(4−1)2+(−6+2)2 = √9+16 = 5
The circle equation is (x−4)2+(y+6)2=52 ⇒ x2 + y2 - 8x + 12y + 27 = 0
Case 2 : a = - 6 ⇒ b = 2(- 6) - 14 = 26 ∴ Centre C = (-6, - 26)
Radius = CP = √(−6−1)2+(−26+2)2 = √49+576 = √625 = 25
The circle equation is (x+6)2+(y+26)2=252 ⇒ x2 + y2 + 12x + 52y + 87 = 0
∴ The circles are x2 + y2 - 8x + 12y + 27 = 0, x2 + y2 + 12x + 52y + 87 = 0
Question 13
If the chord y = mx + 1 of the circle x2+y2=1 subtends an angle of measure 45∘ at the major segment of the circle then value of m is
SOLUTION
Solution : C
x2+y2–(y–mx)2=0
Coefficient x2 +Coefficient y2=0
m=+1 or -1
Question 14
The equation of the image of the circle x2+y2 - 6x - 4y + 12 = 0 by the line mirror x + y - 1 = 0
SOLUTION
Solution : A
For the circle x2+y2 - 6x - 4y + 12 = 0.
This circle has centre C(3, 2) and radius (r) = 1
Image of (3, 2) w.r.t the line x + y - 1 = 0 = (-1, -2)
Equation of the circle with centre (-1, -2) and radius 1 is,
(x+1)2+(y+2)2 = 1
∴ Image of x2+y2 - 6x - 4y + 12 = 0 w.r.t x + y - 1 = 0 is,
x2+y2 + 2x + 4y + 4 = 0
Question 15
If x + y + k = 0 is a tangent to the circle x2+y2 - 2x - 4y + 3 = 0 then k =
SOLUTION
Solution : B
Radius = Distance from the centre C(1, 2) to x + y + k = 0
⇒ √1+4−3 = |1 + 2 + k|√2 ⇒ |k+3| = 2 ⇒ k + 3 = ± 2 ⇒ k = -1, - 5