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Conic Sections Test - 2

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Conic Sections Test - 2

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Question 1
2 / -0.83

The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle of area 154 sq. units. The equation of the circle is

A
x² + y² – 2x – 2y = 47

B
x² + y² – 2x – 2y = 62

C
x² + y² – 2x + 2y = 47

D
x² + y² – 2x + 2y = 62

Solution
Question 2
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If a be the radius of a circle which touches x-axis at the origin, then its equation is

A
x² + y² + ax = 0

B
x² + y² ±2ya = 0

C
x² + y² ±2xa = 0

D
x² + y² + ya = 0

Solution

The equation of the circle with centre at (h,k) and radius equal to a is (x −h)² +(y −k)² = a²
When the circle passes through the origin and centre lies on x −axis
⇒h = a and k = 0
Then the equation (x −h)² +(y −k)² =a² becomes (x −a)² +y² =a²
If a circle passes through the origin and centre lies on x −axis then the abscissa will be equal to the radius of the circle and the y −co-ordinate of the centre will be zero Hence, the equation of the circle will be of the form
(x ±a)² +y² =a² ⇒x² +a² ±2ax+y² =a²
=x² +y² ±2ax=0 is the required equation of the circle.
Question 3
2 / -0.83

The equation of the circle passing through (3, 6) and whose centre is (2, –1) is

A
x² + y² – 4x + 2y = 45

B
x² + y² – 4x – 2y + 45 = 0

C
x² + y² + 4x – 2y = 45

D
x² + y² – 4x + 2y + 45 = 0

Solution

(x −2)² +(y+1)² = r²
(3,6). lies on it
⇒ 1+49=r²
⇒ r² =50
⇒ x² +4 −4x+y² +1+2y=50.
⇒ x² +y² −4x+2y −45=0
Question 4
2 / -0.83

The equation of a circle which passes through the three points (3, 0) (1, –6), (4, –1) is

A
2x² + 2y² + 5x – 11y + 3 = 0

B
x² + y² – 5x + 11y – 3 = 0

C
x² + y² + 5x – 11y + 3 = 0

D
2x² + 2y² – 5x + 11y – 3 = 0

Solution

Gen equation of a circle is (x −h)² +(y −k)² =r² (3,0),(1,−6),(4,−1) circle passes through these points
∴These three points should satisfy the equation
of circle.
(3 −h)² +k² =r² ___ (I)
(1 −h)² +(−6 −k)² = r² (1 −h)² +(6+k)² =r² ___ (II)
From (I) &(II)
(3 −h)² +k² =(1 −h)² +(6+k)²
9+h² −6h+k² =1+h² 2h+36+k² +12k
9 −6h=1 −2h+36+12k
9 −37=4h+12k
−28=4h+12k
h+3k=−7
h=−7 −3k ___ (III)
(4 −h)² +(−1 −k)² =r² ___ (IV)
From (I) &(IV)
(3 −h)² +k² = (4 −h)² +(1+k)²
9+h² −6h+k²
=16+h² −8h+1+k² +2k
9 −6h=17 −8h+2k
9 −17=−2h+2k
−8=−2h+2k
−h+k=−4
k=−4+h ___ (V)
Put (V) in (III)
h=−7 −3(−4+h)
h=−7+12 −3h
h=5/4,k=−11/4,r=170/16
Eq of circle: (x −5/4)² +(y+11/4)² =170/16
x² +25/16 −10x/4+y² +121/16+22y/4=170/16
Simplifying, we get 2x² +xy² −5x+11y −3=0
Question 5
2 / -0.83

y = √3x + c₁ &y = √3x + c₂ are two parallel tangents of a circle of radius 2 units, then |c₁ – c₂ | is equal to

A
8

B
4

C
2

D
1

Solution

For both lines to be parallel tangent the distance between both lines
should be equal to the diameter of the circle
⇒4 = |c₁ −c₂ |/(1+3)^1/2
⇒∣c₁ −c₂ ∣= 8
Question 6
2 / -0.83

B and C are fixed point having co-ordinates (3, 0) and (–3, 0) respectively. If the vertical angle BAC is 90 º, then the locus of the centroid of the DABC has the equation

A
x² + y² = 1

B
x² + y² = 2

C
9(x² + y² ) = 1

D
9(x² + y² ) = 4

Solution

Let A (a, b) and G (x. y) Now A, G, O are collinear
⇒x = (2*0 + a)/3
⇒a = 3x
and similarly b = 3y
Now (a,b) lies on the circle = 9x² + 9y² = 9
=>x² + y² = 1
Question 7
2 / -0.83

The area of an equilateral triangle inscribed in the circle x² + y² – 2x = 0 is

A

B

C

D
None

Solution
Question 8
2 / -0.83

The length of intercept on y-axis, by a circle whose diameter is the line joining the points (–4,3) and (12,–1) is

A
2 √3

B
3 √3

C

D
None of these

Solution

Equation of circle is given by,
Question 9
2 / -0.83

The gradient of the tangent line at the point (a cos a, a sin a) to the circle x² + y² = a² , is

A
tan(p –a)

B
tan a

C
cot a

D
– cot a

Solution
Question 10
2 / -0.83

lx + my + n = 0 is a tangent line to the circle x² + y² = r² , if

A
l² + m² = n² r²

B
l² + m² = n² + r²

C
n² = r² (l² + m² )

D
None of these

Solution
Question 11
2 / -0.83

If y = c is a tangent to the circle x² +y² –2x+2y –2 = 0 at (1, 1), then the value of c is

A
1

B
2

C
–1

D
–2

Solution

For line y=c to be tangent to the given circle at point (1,1)
It has to pass through (1,1)
⇒c = 1
Question 12
2 / -0.83

Line 3x + 4y = 25 touches the circle x² + y² = 25 at the point

A
(4, 3)

B
(3, 4)

C
(–3, –4)

D
None of these

Solution

Equation of tangent at (x1 , y1) to the circle x² + y² =25 is given by,
xx1 + yy1 = 25. Now comparing it with 3x+4y = 25
We get required point of contact (3,4)
Question 13
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The equations of the tangents drawn from the point (0, 1) to the circle x² + y² - 2x + 4y = 0 are

A
2x - y + 1 = 0, x + 2y - 2 = 0

B
2x - y - 1 = 0, x + 2y - 2 = 0

C
2x - y + 1 = 0, x + 2y + 2 = 0

D
2x - y - 1 = 0, x + 2y + 2 = 0

Solution

Let equation of tangent with slope =m and point (0,1)
(y −1)=m(x −0)⇒y=mx+1
Intersection point
x² +(mx+1)² −2x+4(mx+1)=0
(1+m² )x² +(−2+6m)x+5=0
For y=mx+1 to be tangent, discriminant =0
(6m −2)² −4 ×5(1+m² )=0
36m² +4 −24m −20m² +20=0
16m² −20m+24=0
⇒2m² −3m −2=0
(2m+1)(m −2)=0
Question 14
2 / -0.83

The greatest distance of the point P(10, 7) from the circle x² + y² – 4x – 2y – 20 = 0 is

A
5

B
15

C
10

D
None of these

Solution

Centre and radius of the given circle are C(2,1) and √4+1+20 = 5 respectively.
Now CP=√8² +6² =10. Hence greatest distance of point P from the given circle is =10+r=15
Question 15
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The equation of the normal to the circle x² +y² = 9 at the point is

A
x – y = √2/3

B
x + y = 0

C
x – y = 0

D
None of these

Solution

(x-x1)/(x1+g) = (y-y1)/(y1+g)
[x-1/(2)^½ ]/(1/(2)^½ ) = [y-1/(2)^½ ]/(1/(2)^½ )
[(2x)^½ -1 *(2)^½ ]/(2)^½ = [(2x)^½ -1 *(2)^½ ]/(2)^½
(2x)^½ -1 = (2y)^½ -1
= x-y = 0
Question 16
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The parametric coordinates of any point on the circle x² + y² – 4x – 4y = 0 are

A
(–2 + 2cosa, –2 + 2 sin a)

B
(2 + 2cosa, 4 + 2 sina)

C
(2 + 2 √2cosa, 2 + 2 √2sina)

D
None of these

Solution
Question 17
2 / -0.83

The length of the tangent drawn from the point (2, 3) to the circles 2(x² + y² ) – 7x + 9y – 11 = 0.

A
18

B
14

C

D

Solution
Question 18
2 / -0.83

Tangents are drawn from (4, 4) to the circle x² + y² – 2x – 2y – 7 = 0 to meet the circle at A and B. The length of the chord AB is

A
2 √3

B
3 √2

C
2 √6

D
6 √2

Solution
Question 19
2 / -0.83

The angle between the two tangents from the origin to the circle (x – 7)² + (y + 1)² = 25 equals

A
π/2

B
π/3

C
π/4

D
None

Solution

Let tangent from origin be y = mx
Using the condition of tangency, we get

The angle between tangents = π/2
Question 20
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Pair of tangents are drawn from every point on the line 3x + 4y = 12 on the circle x² + y² = 4. Their variable chord of contact always passes through a fixed point whose co-ordinates are

A

B

C
(1, 1)

D

Solution

Let P(x₁ ,y₁ ) be a point on the line 3x + 4y = 12
Equation of variable chord of contact of P(x₁ ,y1₎ w.r.t circle x2 + y2 = 4
xx₁ + yy₁ −4 = 0 ...(1)
Also 3x₁ + 4y1 −12 = 0
⇒x₁ + 4/3y₁ −4 = 0 ...(2)
Comparing (1) and (2), we get
x = 1; y = 4/3
∴Variable chord of contact always passes through (1, 4/3)
Question 21
2 / -0.83

The locus of the mid-points of the chords of the circle x² + y² – 2x – 4y – 11 = 0 which subtend 60 ºat the centre is

A
x² + y² – 4x – 2y – 7 = 0

B
x² + y² + 4x + 2y – 7 = 0

C
x² + y² – 2x – 4y – 7 = 0

D
x² + y² + 2x + 4y + 7 = 0

Solution

Let AB be the chord of the circle and P be the midpoint of AB.
It is known that perpendicular from the center bisects a chord.
Thus △ACP is a right-angled triangle.
Now AC=BC= radius.
The equation of the give circle can be written as
(x −1)² +(y −2)² =16
Hence, centre C=(1,2) and radius =r=4 units.
PC=ACsin60degree
= rsin60degree
= 4([2(3)^½ ]/2
= 2(3)^1/2 units
Therefore, PC=2(3)^1/2
⇒PC² =12
⇒(x −1)² +(y −2)² =12
⇒x² +y² −2x −4y+5=12
⇒x² +y² −2x −4y −7=0
Question 22
2 / -0.83

The locus of the centres of the circles such that the point (2, 3) is the mid point of the chord 5x + 2y = 16 is

A
2x – 5y + 11 = 0

B
2x + 5y – 11 = 0

C
2x + 5y + 11 = 0

D
None

Solution

Slope of the given chord = −5/2
Slope of the line joining the midpoint on the chord and the centre of the circle = (3+f)/(2+g)
(5/2)[(3+f)/(2+g)] = −1
⇒15 + 5f = 4 + 2g
⇒Locus is 2x −5y + 11 = 0
Question 23
2 / -0.83

The equation of the circle having the lines y² – 2y + 4x – 2xy = 0 as its normals &passing through the point (2, 1) is

A
x² + y² – 2x – 4y + 3 = 0

B
x² + y² – 2x + 4y – 5 = 0

C
x² + y² + 2x + 4y – 13 = 0

D
None

Solution

The normal line to circle is →y ²- 2 y + 4 x -2 xy=0
→y(y-2) - 2x(y-2)=0
→(y-2)(y-2x)=0
the two lines are , y=2 and 2 x -y =0
The point of intersection of normals are centre of circle.
→Put , y=2 in 2 x -y=0, we get
→2 x -2=0
→2 x=2
→x=1
So, the point of intersection of normals is (1,2) which is the center of circle.
Also, the circle passes through (2,1).
Radius of circle is given by distance formula = [(1-2)²+ (2-1)²]^½
=(1+1)^½ =(2)^½
The equation of circle having center (1,2) and radius √2 is
= (x-1)²+(y-2)²=[√2]²
→(x-1)²+(y-2)²= 2
x ²+y ²-2x-4y+3 = 0
Question 24
2 / -0.83

A circle is drawn touching the x-axis and centre at the point which is the reflection of (a, b) in the line y – x = 0. The equation of the circle is

A
x² + y² – 2bx – 2ay + a² = 0

B
x² + y² – 2bx – 2ay + b² = 0

C
x² + y² – 2ax – 2by + b² = 0

D
x² + y² – 2ax – 2by + a² = 0

Solution

Radius of the circle = a
and centre ≡(b,a)
And circle is touching the x-axis.
The equation of the circle :
x² +y² −2bx −2ay+b² = 0
Question 25
2 / -0.83

The number of common tangents of the circles x² + y² – 2x – 1 = 0 and x² + y² – 2y – 7 = 0

A
1

B
2

C
3

D
4

Solution

Internally touch ∴common tangent is one.
Question 26
2 / -0.83

The point from which the tangents to the circles x² + y² – 8x + 40 = 0, 5x² + 5y² – 25 x + 80 = 0, x² + y² – 8x + 16y + 160 = 0 are equal in length is

A

B

C

D
None of these

Solution

The Required point is the radical centre of the three given circles. The radical axes of the three circles taken in pairs are 3x - 24 = 0,16y + 120 = 0 and - 3x + 16y + 80 = 0. On solving, the required point is (8, -15/2).
Question 27
2 / -0.83

If the circle x² + y² = 9 touches the circle x² + y² + 6y + c = 0, then c is equal to

A
27

B
36

C
–36

D
–27

Solution

touches the another circle

Now, Central first circle will be
And its radius will be 3 units.
Also,centre of second circle
And radius,

As both touches each other
So,
Question 28
2 / -0.83

The tangent from the point of intersection of the lines 2x – 3y + 1 = 0 and 3x – 2y –1 = 0 to the circle x² + y² + 2x – 4y = 0 is

A
x + 2y = 0, x – 2y + 1 = 0

B
2x – y – 1 = 0

C
y = x, y = 3x – 2

D
2x + y + 1 = 0

Solution
Question 29
2 / -0.83

The length of the common chord of circles x² + y² – 6x – 16 = 0 and x² + y² – 8y – 9 = 0 is

A

B

C
5/2

D
none of these

Solution
Question 30
2 / -0.83

The distance between the chords of contact of tangents to the circle x² + y² + 2gx + 2fy + c = 0 from the origin and from the point (g, f) is

A

B

C

D

Solution

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