# Question-1

**Find the centre and radius of the following circles:**

(i) x

(ii) x

(iii) x

(iv) 3x

(v) (x - 3)(x - 5) + (y - 7)(y - 1) = 0(i) x

^{2}+ y^{2}= 1(ii) x

^{2}+ y^{2}â€“ 4x â€“ 6y â€“ 9 = 0(iii) x

^{2}+ y^{2}- 8x - 6y - 24 = 0(iv) 3x

^{2}+ 3y^{2}+ 4x - 4y - 4 = 0(v) (x - 3)(x - 5) + (y - 7)(y - 1) = 0

**Solution:**

(i)

**The general equation of circle is x**

^{2}+ y

^{2}+ 2gx + 2fy + c = 0.

2g = 0, 2f = 0, c = -1

âˆ´ centre is (-g, -f) = (0, 0)

Radius = == 1units

(ii)

**The general equation of circle is x**

^{2}+ y

^{2}â€“ 4x â€“ 6y â€“ 9 = 0.

2g = -4, 2f = -6, c = -9

g = -2, f = -3

âˆ´ centre is (-g, -f) = (2, 3)

Radius = = units

(iii)

**The general equation of circle is x**

^{2}+ y

^{2}- 8x - 6y - 24 = 0.

2g = -8, 2f = -6, c = -24

g = -4, f = -3

âˆ´ centre is (-g, -f) = (4, 3)

Radius = = = 1units

(iv)

**The general equation of circle is 3x**

^{2}+ 3y

^{2}+ 4x - 4y - 4 = 0.

2g = 4/3, 2f = -4/3, c = -4/3

âˆ´ centre is (-g, -f) = (-2/3, 2/3)

Radius = = = = unit

(v) (x - 3)(x - 5) + (y - 7)(y - 1) = 0

x

^{2}â€“ 8x + 15 + y

^{2}â€“ 8y + 7 = 0

The general equation of circle is x

^{2}+ y

^{2}â€“ 8x â€“ 8y + 22 = 0.

2g = -8, 2f = -8, c = 22

âˆ´ centre is (-g, -f) = (4, 4)

Radius = == = âˆš 10 units

# Question-2

**For what values of a and b does the equation (a - 2)x**

^{2}+ by^{2}+ (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 x

^{2}+ y

^{2}+ 2gx + 2fy + c = 0.

2fy + c = 0.

The equation (a - 2)x

^{2 }+ by

^{2}+ (b - 2)xy + 4x + 4y - 1 = 0 when compared with the general equation we get,

Coefficient of x

^{2 }= Coefficient of y

^{2}

Thus, a - 2 = b â€¦â€¦â€¦â€¦â€¦â€¦.(i)

Coefficient of xy is zero (i.e.) h = 0

b â€“ 2 = 0

b = 2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

Substitute (ii) in (i)

a = b + 2 = 2 + 2 = 4

âˆ´ The required equation of circle is 2x

^{2}+ by

^{2}+ 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 â€“ k)

^{2}= r

^{2 }Here (h, k) = (1, 2)

(x - 1)

^{2}+ (y â€“ 2)

^{2}= r

^{2 }is the required equation of the circle.

The circle passes through (2, 3).

âˆ´ (2 - 1)

^{2}+ (3 â€“ 2)

^{2}= r

^{2}âˆ´ (1)

^{2}+ (1)

^{2}= r

^{2}âˆ´ r

^{2}= 2

âˆ´ The required equation of the circle is (x - 1)

^{2}+ (y â€“ 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 â€¦â€¦â€¦..(i)

2x + y = 8 â€¦â€¦â€¦..(ii)

(i) Ã— 2 â€“ (ii)

2x + 4y = 14

3y = 6

y = 2

âˆ´ x = 7 â€“ 2y = 7 â€“ 2(2) = 7 â€“ 4 = 3

The two diameters of a circle intersect each other at a point which is the centre of the circle.

âˆ´ (3, 2) is the centre of a circle.

âˆ´ The equation of the circle (x - 3)

^{2}+ (y - 2)

^{2}= 5

^{2}.

# Question-5

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

**Solution:**

Area of a circle = 16Ï€ square unitsÏ€ r

^{2 }= 16Ï€ r

^{2 }= 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 8Ï€ units.**

**Solution:**

Circumference of a circle = 8Ï€ units

2Ï€ r

^{ }= 16Ï€ 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 x**

^{2}+ y^{2}â€“ 6x - 8y + 15 = 0.**Solution:**

x

^{2}+ y

^{2}â€“ 6x - 8y + 15 = 0

2g = -6, 2f = -8, c = 15

g = -3, f = -4

r

^{2}= g

^{2}+ f

^{2}â€“ c = 9 + 16 â€“ 15 = 10

Area of the circle = Ï€ r

^{2}= 10Ï€ square units

Circumference of the circle = 2 Ï€ r = Ï€ 2âˆš 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.

âˆ´ The equation of the circle is (x - h)

^{2}+ y

^{2}= r

^{2}.

(2 - h)

^{2}+ 3

^{2}= 5

^{2}

4 â€“ 4h + h

^{2}+ 9 = 25

h

^{2}â€“ 4h - 12 = 0

(h â€“ 6)(h + 2) = 0

h = 6, -2

(x - 6)

^{2}+ y

^{2}= 25 or (x + 2)

^{2}+ y

^{2}= 25

x

^{2}â€“ 12x + 36 + y

^{2}= 25 or x

^{2}+ 4x + 4 + y

^{2}= 25

x

^{2}â€“ 12x + y

^{2}+ 11 = 0 or x

^{2}+ 4x + y

^{2}â€“ 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 â€“ x

_{1})(x â€“ x

_{2}) + (y â€“ y

_{1})(y â€“ y

_{2}) = 0.

Here (x

_{1}, y

_{1}) = (1, 2) and (x

_{2}, y

_{2}) = (2, 4)

âˆ´ (x â€“ 1)(x â€“ 2) + (y â€“ 2)(y â€“ 4) = 0.

x

^{2}+ y

^{2 }- 3x â€“ 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 x

^{2}+ y

^{2}+ 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,

âˆ´ 1 + 2g + c = 0

2g + c = 1 â€¦â€¦â€¦â€¦â€¦..(i)

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

1â€“ 2f + c = 0

-2f + c = -1 â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

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

1 + 2f + c = 0

2f + c = -1 â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

(ii) + (iii)

2c = -2

c = -1

Substituting c = -1 in (i)

2g â€“ 1 = 1

g = 1

Substituting c = -1 in (ii)

-2f â€“ 1 = -1

f = 0

The general equation of circle is x

^{2}+ y

^{2}+ 2x - 1 = 0.

# Question-11

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

(3, 2).

(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.

âˆ´ 1 + 1 + 2g + 2f + c = 0

2g + 2f + c = -2 â€¦â€¦â€¦â€¦â€¦..(i)

4 + 1 + 4g - 2f + c = 0

4g - 2f + c = -5 â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

9 + 4 + 6g + 4f + c = 0

6g + 4f + c = -13 â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

(i) - (ii)

-2g + 4f = 3 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iv)

(i) â€“ (iii)

-4g - 2f = 11 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(v)

(iv) â€“ [2(v)]

-10g = 25

Substitute g = in (iv)

-2g + 4f = 3

5 + 4f = 3

4f = 3 â€“ 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,

âˆ´ 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 x

^{2}+ y

^{2}+ 2gx + 2fy + c = 0

(4, 1) lies on the circle.

16 + 1 + 8g + 2f + c = 0

8g + 2f + c = -17 â€¦â€¦â€¦â€¦â€¦â€¦(i)

(6, 5) lies on the circle.

36 + 25 + 12g + 10f + c = 0

12g + 10f + c = -61 .â€¦â€¦â€¦â€¦â€¦(ii)

(ii) â€“ (i)

4g + 8f = -44

g + 2f = -11 .â€¦â€¦â€¦â€¦â€¦(iii)

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

âˆ´ -4g - f = 16 .â€¦â€¦â€¦â€¦â€¦(iv)

2(iv) + (iii)

-7g = 21

g = -3

Substitute g = -3 in (iv)

-4(-3) â€“ f = 16

f = -4

Substitute in (i)

8(-3) + 2(-4) + c = -17

-24 â€“ 8 + c = -17

c = -17 + 32

c = 15 âˆ´ The general equation of the circle is x^{2} + y^{2} - 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 x

^{2}+ y

^{2}+ 2gx + 2fy + c = 0. (-g, -f) is centre of the circle.

Centre (-g, -f) lies on the line x

**= 2y.**

âˆ´ -g = -2f

g = 2f â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

(-1, 2) lies on the circle.

âˆ´ 1 + 4 â€“ 2g + 4f + c = 0

1 + 4 â€“ 2(2f) + 4f + c = 0 (from i)

âˆ´ c = -5 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

(3, -2) lies on the circle.

âˆ´ 9 + 4 + 6g - 4f + c = 0

âˆ´ 9 + 4 + 6(2f) - 4f - 5 = 0 (from i and ii)

8f + 8 = 0

f = -1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iii)

g = -2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iv)âˆ´ The required equation of the circle is x^{2} + y^{2} - 4x - 2y - 5 = 0.

# Question-14

**Find the cartesian equation of the circle whose parametric equations are x = cosÎ¸ , y = sin Î¸ and 0 â‰¤ Î¸ â‰¤ 2Ï€ .**

**Solution:**

cosÎ¸ = 4x

sinÎ¸ = 4y

cos

^{2}Î¸

^{ }+ sin

^{2}Î¸

^{ }= 1

âˆ´ 16x

^{2}+ 16y

^{2 }= 1 is the required cartesian equation of the circle.

# Question-15

**Find the parametric equation of the circle 4x**

^{2}+ 4y^{2}= 9.**Solution:**

r

^{2}= 9/4

\ r = 3/2

âˆ´ The parametric equations of the given circle 4x

^{2}+ 4y

^{2}= 9 are

x = cosÎ¸ and y = sinÎ¸ , 0 â‰¤ Î¸ â‰¤ 2Ï€ .

# Question-16

**Find the coordinates of foci, equations of the directrices, and the length of the latusrectum of the parabola: y**^{2}= 12x.**Solution:**

The given problem y

^{2}= 12x is of the form y

^{2}= 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 = -4x**^{2}+3x.**Solution:**

y = -4x

^{2 }+ 3x

4x

^{2 }- 3x = -y

x

^{2 }- x =

x

^{2 }- 2x+ = +

=

=

^{ }= 4 xâ€¦â€¦.. (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 Ãž x =

Y = 0

=0 Ãž 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 Ãž 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

x

^{2}= (x - 6)

^{2}+ y

^{2}

x

^{2 }= x

^{2 }- 12x + 36 + y

^{2}

y

^{2 }- 12x + 36 = 0

y

^{2}= 12x â€“ 36

# Question-19

**Does the point (7, -11) lie inside or outside the circle x**

^{2}+ y^{2}â€“ 10x = 0?**Solution:**

By substituting the point (7, -11) in the equation x

^{2}+ y

^{2}â€“ 10x, we get

7

^{2}+ (-11)

^{2}â€“ 10(7) = 49 + 121 â€“ 70

= 170 â€“ 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 x**

^{2 }+ y^{2}â€“ 5x + 2y â€“ 5 = 0.**Solution:**

By substituting (-2, 1) in the equation x2 + y2 â€“ 5x + 2y â€“ 5 we get,

(-2)2 + (1)2 â€“ 5(-2) + 2(1) â€“ 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 â€“ 5x + 2y â€“ 5 we get,

(0)2 + (0)2 â€“ 5(0) + 2(0) â€“ 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 â€“ 5x + 2y â€“ 5 we get,

(4)2 + (-3)2 â€“ 5(4) + 2(-3) â€“ 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:**

3x

^{2}+ 2y

^{2 }= 6.

Dividing by 6 we get,

This equation is of the form where a^{2} = 2 and b^{2} = 3.(i.e.,) a = and b =

Here a < b, so the major and minor axes of the given ellipse are along y and x â€“ 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 â€“minor axis is 3.****Solution:**

Let S and Sâ€™ be two foci of the required ellipse. Then the coordinates of S and Sâ€™ are (4, 3) and (-4, 3) respectively.

SSâ€™ = 8

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

Then SSâ€™ = 2ae, but 2ae = 8

Thus, ae = 4.

Now, b^{2} = a^{2}(1 â€“ e^{2})

9 = a^{2} - 4^{2}

9 + 16 = a^{2}

a = 5

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

Then,

SP + Sâ€™P = 2 a

9x^{2} + 25y^{2 }- 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}= 6

^{2}.

x

^{2 }+ y

^{2}â€“ 10x â€“ 12y + 25 + 36 = 36

x

^{2 }+ y

^{2}â€“ 10x â€“ 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} = 5^{2}.

x^{2 }+ y ^{2} â€“ 10x â€“ 12y + 36 = 0