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Question-1

Find the centre and radius of the following circles:
(i) x2 + y2 = 1
(ii) x2 + y2 – 4x – 6y – 9 = 0
(iii) x2 + y2 - 8x - 6y - 24 = 0
(iv) 3x2 + 3y2 + 4x - 4y - 4 = 0
(v) (x - 3)(x - 5) + (y - 7)(y - 1) = 0

Solution:
(i) The general equation of circle is x2 + y2 + 2gx + 2fy + c = 0.
2g = 0, 2f = 0, c = -1
centre is (-g, -f) = (0, 0)
  Radius = == 1units

(ii) The general equation of circle is x2 + y2 – 4x – 6y – 9 = 0.
2g = -4, 2f = -6, c = -9
  g = -2, f = -3
centre is (-g, -f) = (2, 3)
 Radius = = units

(iii) The general equation of circle is x2 + y2 - 8x - 6y - 24 = 0.
2g = -8, 2f = -6, c = -24
g = -4, f = -3
centre is (-g, -f) = (4, 3)
 Radius = = = 1units

(iv) The general equation of circle is 3x2 + 3y2 + 4x - 4y - 4 = 0.
2g = 4/3, 2f = -4/3, c = -4/3
centre is (-g, -f) = (-2/3, 2/3)
 Radius = = = = unit

(v) (x - 3)(x - 5) + (y - 7)(y - 1) = 0
x2 – 8x + 15 + y2 – 8y + 7 = 0
The general equation of circle is x2 + y2 – 8x – 8y + 22 = 0.
2g = -8, 2f = -8, c = 22
centre is (-g, -f) = (4, 4)
Radius = == = 10 units

Question-2

For what values of a and b does the equation (a - 2)x2 + by2 + (b - 2)xy + 4x + 4y - 1 = 0 represent a circle? Write down the resulting equation of the circle.

Solution:
The general condition for a second degree equation to represent a circle is x2 + y2 + 2gx + 2fy + c = 0.
2fy + c = 0.
The equation (a - 2)x2 + by2 + (b - 2)xy + 4x + 4y - 1 = 0 when compared with the general equation we get,
Coefficient of x2 = Coefficient of y2
Thus, a - 2 = b ……………….(i)
Coefficient of xy is zero (i.e.) h = 0
b – 2 = 0
b = 2 …………………(ii)
Substitute (ii) in (i)
a = b + 2 = 2 + 2 = 4
The required equation of circle is 2x2 + by2 + 4x + 4y - 1 = 0.

Question-3

Find the equation of the circle passing through the point (2,3) and having its centre at (1, 2).

Solution:
If the centre is {h, k) and radius is r, then the equation of the circle is
(x - h)2 + (y – k)2 = r2
           
Here (h, k) = (1, 2)
(x - 1)2 + (y – 2)2 = r2 is the required equation of the circle.
The circle passes through (2, 3).
(2 - 1)2 + (3 – 2)2 = r2          (1)2 + (1)2 = r2                       r2 = 2
The required equation of the circle is (x - 1)2 + (y – 2)2 = 2.

Question-4

x + 2y = 7, 2x + y = 8 are two diameters of a circle with radius 5 units. Find the equation of the circle.

Solution:
x + 2y = 7 ………..(i)
2x + y = 8 ………..(ii)
(i) × 2 – (ii)
2x + 4y = 14
3y = 6
y = 2
x = 7 – 2y = 7 – 2(2) = 7 – 4 = 3
The two diameters of a circle intersect each other at a point which is the centre of the circle.
(3, 2) is the centre of a circle.
The equation of the circle (x - 3)2 + (y - 2)2 = 52.

Question-5

The area of a circle is 16π square units. If the centre of the circle is (7, - 3), find the equation of the circle.

Solution:
Area of a circle = 16π square unitsπ r2 = 16π     r2 = 16
The equation of the circle is (x - 7)2 + (y + 3)2 = 16.

Question-6

Find the equation of the circle whose centre is (- 4, 5) and circumference is 8π units.

Solution:
Circumference of a circle = 8π units
2π r = 16π    2r = 16
r = 8 units
The equation of the circle is (x + 4)2 + (y - 5)2 = 64.

Question-7

Find the circumference and area of the circle x2 + y2 – 6x - 8y + 15 = 0.

Solution:
x2 + y2 – 6x - 8y + 15 = 0
2g = -6, 2f = -8, c = 15
g = -3, f = -4
r2 = g2 + f2 – c = 9 + 16 – 15 = 10
Area of the circle = π r2 = 10π square units
Circumference of the circle = 2 π r = π 2 10 units

Question-8

Find the equation of the circle which passes through (2, 3) and whose centre is on x-axis and radius is 5 units.

Solution:
Centre is on x-axis.
The equation of the circle is (x - h)2 + y2 = r2.
(2 - h)2 + 32 = 52
4 – 4h + h2 + 9 = 25
h2 – 4h - 12 = 0
(h – 6)(h + 2) = 0
h = 6, -2
(x - 6)2 + y2 = 25           or     (x + 2)2 + y2 = 25
x2 – 12x + 36 + y2 = 25 or      x2 + 4x + 4 + y2 = 25
x2 – 12x + y2 + 11 = 0    or      x2 + 4x + y2 – 21 = 0 are the equations of the circle.

Question-9

Find the equation of the circle described on the line joining the points (1, 2) and (2, 4) as its diameter.

Solution:
The equation of the circle is (x – x1)(x – x2) + (y – y1)(y – y2) = 0.
Here (x1, y1) = (1, 2) and (x2, y2) = (2, 4)
  (x – 1)(x – 2) + (y – 2)(y – 4) = 0.
                x2 + y2 - 3x – 6y + 10 = 0.

Question-10

Find the equation of the circle passing through the points (1, 0), (0, - 1) and (0,1).

Solution:
The general equation of circle is x2 + y2 + 2gx + 2fy + c = 0.
The points (1, 0), (0, - 1) and (0,1) lie on the circle.

If (1, 0) lie on the circle x2 + y2 + 2gx + 2fy + c = 0, we get,
1 + 2g + c = 0
         2g + c = 1 ……………..(i)

If (0, -1) lie on the circle x2 + y2 + 2gx + 2fy + c = 0, we get,
1– 2f + c = 0

-2f + c = -1 ……………….(ii)

If (0, 1) lie on the circle x2 + y2 + 2gx + 2fy + c = 0, we get,
1 + 2f + c = 0

2f + c = -1 ………………..(iii)

(ii) + (iii)
2c = -2
  c = -1
Substituting c = -1 in (i)
2g – 1 = 1
       g = 1
Substituting c = -1 in (ii)
-2f – 1 = -1
        f = 0
The general equation of circle is x2 + y2 + 2x - 1 = 0.

Question-11

Find the equation of the circle passing through the points (1, 1), (2, -1) and
(3, 2).

Solution:
The general equation of circle is x2 + y2 + 2gx + 2fy + c = 0.
The points (1, 1), (2, -1) and (3, 2) lie on the circle.
1 + 1 + 2g + 2f + c = 0
2g + 2f + c = -2 ……………..(i)

4 + 1 + 4g - 2f + c = 0
4g - 2f + c = -5 ……………….(ii)

9 + 4 + 6g + 4f + c = 0
6g + 4f + c = -13 ………………..(iii)

(i) - (ii)
-2g + 4f = 3 ………………….(iv)

(i) – (iii)
-4g - 2f = 11 ………………….(v)

(iv) – [2(v)]
-10g = 25

Substitute g = in (iv)
-2g + 4f = 3

5 + 4f = 3
4f = 3 – 5
    = -2
  f =
Substitute g = and f = in (i)
2g + 2f + c = -2
2 + 2 + c = -2
-5 +(- 1) + c = -2
c = -2 + 6 = 4
By substituting g = , f = and c = 4 in x2 + y2 + 2gx + 2fy + c = 0 we get,
The general equation of circle is x2 + y2 - 5x - y + 4 = 0.

Question-12

Find the equation of the circle that passes through the points (4, 1) and (6, 5) and has its centre on the line 4x + y = 16.

Solution:
The general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0
(4, 1) lies on the circle.
16 + 1 + 8g + 2f + c = 0
            8g + 2f + c = -17 ………………(i)

(6, 5) lies on the circle.
36 + 25 + 12g + 10f + c = 0
               12g + 10f + c = -61 .……………(ii)

(ii) – (i)
4g + 8f = -44
  g + 2f = -11 .……………(iii)

Let (-g, -f) be the centre of the circle lying on 4x + y = 16.
-4g - f = 16 .……………(iv)

2(iv) + (iii)
-7g = 21
   g = -3
Substitute g = -3 in (iv)
    -4(-3) – f = 16
                f = -4
Substitute in (i)
8(-3) + 2(-4) + c = -17
-24 – 8 + c = -17
c = -17 + 32
c = 15 The general equation of the circle is x2 + y2 - 6x - 8y + 15 = 0.

Question-13

Find the equation of the circle whose centre is on the line x = 2y and which passes through the points (- 1, 2) and (3, -2).

Solution:
The general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0. (-g, -f) is centre of the circle.
Centre (-g, -f) lies on the line x = 2y.
-g = -2f
     g = 2f ………………….(i)

(-1, 2) lies on the circle.
1 + 4 – 2g + 4f + c = 0
1 + 4 – 2(2f) + 4f + c = 0 (from i)
c = -5 …………………….(ii)

(3, -2) lies on the circle.
9 + 4 + 6g - 4f + c = 0
9 + 4 + 6(2f) - 4f - 5 = 0 (from i and ii)
8f + 8 = 0
f = -1 …………………….(iii)
g = -2 …………………….(iv) The required equation of the circle is x2 + y2 - 4x - 2y - 5 = 0.

Question-14

Find the cartesian equation of the circle whose parametric equations are x = cosθ , y = sin θ and 0 θ 2π .

Solution:
cosθ = 4x
   sinθ = 4y
cos2θ + sin2θ = 1
16x2 + 16y2 = 1 is the required cartesian equation of the circle.

Question-15

Find the parametric equation of the circle 4x2 + 4y2 = 9.

Solution:
r2 = 9/4
  \ r = 3/2
The parametric equations of the given circle 4x2 + 4y2 = 9 are
x = cosθ and y = sinθ , 0 θ 2π .

Question-16

Find the coordinates of foci, equations of the directrices, and the length of the latusrectum of the parabola: y2 = 12x.

Solution:
The given problem y2 = 12x is of the form y2 = 4ax, where 4a = 12. (i.e.) a = 3. The coordinates of the focus (a, 0) is (3, 0). The equation of the directrix is x = -a (i.e.) (-3, 0) Length of the latusrectum = 4a = 12.

Question-17

Find the foci, vertices of the parabola: y = -4x2+3x.

Solution:
y = -4x2 + 3x
4x2 - 3x = -y
x2 - x =
x2 - 2x
+ = +
=
            =

= 4 x…….. (i)

 

Shifting the origin to the point without rotating the axes and denoting the new coordinates w.r.t. these axes by X and Y we have,

Put X =

Y =
In original coordinates :vertex is X =0

=0 Þ x =
Y = 0

=0 Þ y =

... Vertex is
Focus: The coordinates of the focus with respect to the new axes are (X = 0, Y = a)
In original coordinates,
Focus =0 Þ x =

=

y = =

... Focus is

Question-18

Find the equation of the parabola whose directrix is x = 0 and focus at (6, 0).

Solution:
Let P(x, y) be a point on the parabola. Join SP.
SP =   =

PM = x
SP = PM

x2 = (x - 6)2 + y2
x2 = x2 - 12x + 36 + y2
y2 - 12x + 36 = 0
y2 = 12x – 36

Question-19

Does the point (7, -11) lie inside or outside the circle x2 + y2 – 10x = 0?

Solution:
By substituting the point (7, -11) in the equation x2 + y2 – 10x, we get
72 + (-11)2 – 10(7) = 49 + 121 – 70
= 170 – 70 = 100 > 0
Thus the point (7, -11) lies outside the circle.

Question-20

Determine whether the points (-2, 1), (0, 0) and (4, -3) lie outside, on or inside the circle x2 + y2 – 5x + 2y – 5 = 0.

Solution:
By substituting (-2, 1) in the equation x2 + y2 – 5x + 2y – 5 we get,
 (-2)2 + (1)2 – 5(-2) + 2(1) – 5 = 4 + 1 + 10 + 2 - 5
                                            = 12 > 0.
Thus the point (0, 0) lie on the circle .
By substituting (0, 0) in the equation x2 + y2 – 5x + 2y – 5 we get,
(0)2 + (0)2 – 5(0) + 2(0) – 5 = 0 + 0 + 0 + 2 - 5
                                        = -5 < 0.
Thus the point (0, 0) lie inside the circle.
By substituting (4, -3) in the equation x2 + y2 – 5x + 2y – 5 we get,
(4)2 + (-3)2 – 5(4) + 2(-3) – 5 = 16 + 9 + 20 - 6 - 5
                                           = 34 > 0.
Thus the point (4, -3) lie on the circle.

Question-21

For the following ellipses find the lengths of major and minor axes, coordinates of foci and vertices and eccentricity 3x2 + 2y2 = 6.

Solution:
3x2 + 2y2 = 6.

 

Dividing by 6 we get,

This equation is of the form where a2 = 2 and b2 = 3.(i.e.,) a = and b =

Here a < b, so the major and minor axes of the given ellipse are along y and x – axes respectively.

Length of the major axis = 2b = 2, Length of the minor axis = 2a = 2.

The coordinates of the vertices are (0, b) and (0, -b).

(i.e.,) (0, ) and (0, -)

The eccentricity e of the ellipse is given by e = .

The coordinates of the foci are (0, be) and (0, -be) (i.e.,) (0, 1) and ( 0, -1).

Question-22

The foci of an ellipse are and its eccentricity is , find its equation.

Solution:
Let the equation of the ellipse be . The coordinates of foci are .

ae = 8

 

Thus

Hence, the equation of the ellipse is

Question-23

Find the equation of the ellipse whose foci are (4, 3), (-4, 3) and whose semi –minor axis is 3.

Solution:
Let S and S’ be two foci of the required ellipse. Then the coordinates of S and S’ are (4, 3) and (-4, 3) respectively.

 

SS’ = 8

Let 2a and 2b be the lengths of the axes of the ellipse and e be the eccentricity.

Then SS’ = 2ae, but 2ae = 8

Thus, ae = 4.

Now, b2 = a2(1 – e2)

9 = a2 - 42

9 + 16 = a2

a = 5

 

Let P(x, y) be any point on the ellipse.

Then,

SP + S’P = 2 a

 

9x2 + 25y2 - 150 y = 0.

Question-24


Solution:
(i) Let P be the point on x-axis where it touches the circle.
Centre C is (5, 6) and P is (5, 0).
r = CP = = 6.
The equation of the circle is (x - 5)2 + (y - 6)2 = 62.
                     x2 + y 2 – 10x – 12y + 25 + 36 = 36
                             x2 + y 2 – 10x – 12y + 25 = 0

(ii) Let P be the point on y-axis where it touches the circle.
Centre C is (5, 6) and P is (0, 6).
r = CP = = 5.
The equation of the circle is (x - 5)2 + (y - 6)2 = 52.
x2 + y 2 – 10x – 12y + 36 = 0





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