Class 11 Applied Mathematics

Circles &
Parabola

Exploring the beauty of conic sections. From the perfect symmetry of circles to the directional focus of parabolas.

The Circle

Symmetry in Motion

ℹ️

Standard Equation

A circle is the locus of a point which moves such that its distance from a fixed point (center) is constant (radius).\n\n

(x-h)^2 + (y-k)^2 = r^2

\n\nAt origin

(0,0)

x^2 + y^2 = r^2

Diameter & General Form

Diameter Form

(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0

Center Calculation

For $x^2 + y^2 + 2gx + 2fy + c = 0$:

Center:

(-g, -f)

, Radius:

\sqrt{g^2 + f^2 - c}

Circle Explorer

Visualize standard and general forms of a circle.

C(0, 0)

r = 5

Scale: 1 unit = 20px

Center h: 0

Center k: 0

Radius r: 5

Standard Form

(x - 0)^2 + (y - 0)^2 = 5^2

General Form ($2gx, 2fy$)

x^2 + y^2 +0x +0y -25.0 = 0

g = 0, f = 0

c = -25.0

Existence ConditionFor a real circle,

g^2 + f^2 - c > 0

. Current:

0^2 + 0^2 - (-25.0) = 25

(Valid)

Wait! Radius Check

The quantity $g^2 + f^2 - c$ determines the nature of the circle:

> 0

Real Circle

= 0

Point Circle

< 0

Imaginary

Tip: g = half of coeff of x, f = half of coeff of y

The Parabola

Parabola Lab

Explore standard forms and their geometric properties.

Focus(2, 0)

Directrixx = -2

Parameter (a): 2

AxisX-axis

LR Length4a = 8

Active Equation

y^2 = 4(2)x

Standard DefinitionDistance from Point to Focus = Distance from Point to Directrix.

Standard Forms Comparison

Four orientations of the standard parabola with vertex at $(0,0)$.

Equation	Focus	Directrix	Latus Rectum	Axis
$y^2 = 4ax$	$(a, 0)$	$x = -a$	$4a$	x-axis
$y^2 = -4ax$	$(-a, 0)$	$x = a$	$4a$	x-axis
$x^2 = 4ay$	$(0, a)$	$y = -a$	$4a$	y-axis
$x^2 = -4ay$	$(0, -a)$	$y = a$	$4a$	y-axis

Practice Laboratory

Conics Workouts

8 Important Problems

CBSE Exam Type

Concept Master

Find the equation of a circle whose center is (2, -3) and radius is 5.

✓ Show Solution

1. Given:

(h, k) = (2, -3), r = 5

2. Formula:

(x-h)^2 + (y-k)^2 = r^2

3. Substituting:

(x-2)^2 + (y+3)^2 = 25

4. Expanding:

x^2 + y^2 - 4x + 6y + 4 + 9 = 25

5. Result:

x^2 + y^2 - 4x + 6y - 12 = 0

EASY

Find the center and radius of the circle

x^2 + y^2 - 8x + 10y - 12 = 0

✓ Show Solution

1. General form:

x^2 + y^2 + 2gx + 2fy + c = 0

2. Comparing:

2g = -8 \implies g = -4

2f = 10 \implies f = 5

c = -12

3. Center:

(-g, -f) = (4, -5)

4. Radius:

r = \sqrt{g^2 + f^2 - c} = \sqrt{(-4)^2 + 5^2 - (-12)} = \sqrt{16 + 25 + 12} = \sqrt{53}

units

EASY

Find the equation of the circle passing through the points (3, 4) and (2, 1) as endpoints of a diameter.

✓ Show Solution

1. Diameter Form:

(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0

2. Substitute

(3, 4)

and

(2, 1)

(x-3)(x-2) + (y-4)(y-1) = 0

x^2 - 5x + 6 + y^2 - 5y + 4 = 0

5. Final:

x^2 + y^2 - 5x - 5y + 10 = 0

MEDIUM

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

✓ Show Solution

1. Let center be

(h, k)

. Radius

r^2 = (4-h)^2 + (1-k)^2 = (6-h)^2 + (5-k)^2

2. Expand & Solve for

h, k

16-8h+h^2+1-2k+k^2 = 36-12h+h^2+25-10k+k^2

4h + 8k = 44 \implies h + 2k = 11

4. Given center on

4x + y = 16 \implies 4h + k = 16

5. Solving simultaneously:

h=3, k=4

. Radius

r^2 = (4-3)^2 + (1-4)^2 = 1 + 9 = 10

6. Equation:

(x-3)^2 + (y-4)^2 = 10 \implies x^2 + y^2 - 6x - 8y + 15 = 0

HARD

Find the focus, axis and the length of the latus rectum for

y^2 = 12x

✓ Show Solution

1. Compare with

y^2 = 4ax \implies 4a = 12 \implies a = 3

2. Focus:

(a, 0) = (3, 0)

3. Axis: x-axis (

y=0

)

4. Length of Latus Rectum:

4a = 12

units

5. Directrix:

x = -3

EASY

Find the equation of the parabola whose focus is (0, -3) and directrix is y = 3.

✓ Show Solution

1. Focus is on y-axis, below origin. This is of the form

x^2 = -4ay

a = 3

(distance from vertex (0,0) to focus (0,-3)).

3. Equation:

x^2 = -4(3)y

4. Result:

x^2 = -12y

MEDIUM

Find the equation of the parabola with vertex at origin, axis along x-axis and passing through (2, 3).

✓ Show Solution

1. Axis along x-axis

\implies y^2 = 4ax

2. Passes through (2, 3)

\implies 3^2 = 4a(2) \implies 9 = 8a \implies a = 9/8

3. Equation:

y^2 = 4(9/8)x \implies y^2 = 4.5x

2y^2 = 9x

MEDIUM

Find the area of the triangle formed by the lines joining the vertex of the parabola

x^2=12y

to the ends of its latus rectum.

✓ Show Solution

1. Parabola

x^2 = 12y \implies 4a = 12 \implies a = 3

2. Vertex

O(0, 0)

. Focus

S(0, 3)

3. Latus Rectum ends:

x^2 = 4ay \implies x = \pm 2a = \pm 6

. Points are

(-6, 3)

and

(6, 3)

4. Base of triangle = distance between

(-6, 3)

and

(6, 3) = 12

units.

5. Height of triangle = distance from

O(0,0)

to focal line

y=3

3

units.

6. Area =

\frac{1}{2} \times 12 \times 3 = 18

sq. units.

HARD

Competitive Corner

Board & Previous Year Questions

CBSE 2023

Find the equation of the circle which touches both the axes and whose radius is 4 units.

Detailed Solution1. Since it touches both axes, center is

(\pm 4, \pm 4)

. 2. There are 4 such circles. Case 1 (Quadrant I): center

(4, 4), r=4

3. Equation:

(x-4)^2 + (y-4)^2 = 16 \implies x^2 + y^2 - 8x - 8y + 16 = 0

4. In general:

x^2 + y^2 \pm 8x \pm 8y + 16 = 0

CBSE 2021

Find the equation of the parabola whose vertex is (0,0) and focus is (-2, 0). Also find directrix.

Detailed Solution1. Vertex (0,0), Focus (-2, 0)

\implies

Opens left

\implies y^2 = -4ax

. 2.

a = 2

. 3. Equation:

y^2 = -8x

. 4. Directrix:

x = a \implies x = 2

Common Board Problem

Find the coordinates of the focus and the length of the latus rectum for the parabola $3x^2 = -8y$.

Detailed Solution1. Rewrite:

x^2 = -\frac{8}{3}y

2. Compare with

x^2 = -4ay \implies 4a = 8/3 \implies a = 2/3

3. Focus:

(0, -a) = (0, -2/3)

4. Latus Rectum length:

4a = 8/3

units.

Concept MCQ

If the equation $x^2 + y^2 + 2gx + 2fy + c = 0$ represents a circle, find the condition that the circle passes through the origin.

Detailed Solution1. Circle passes through

(0, 0)

. 2. Substitute

x=0, y=0

in the equation:

0^2 + 0^2 + 2g(0) + 2f(0) + c = 0

3. Result:

c = 0

Most Repeated

Find the equation of the circle which is concentric with $x^2+y^2-4x-6y-9=0$ and passes through ( -4, -5).

Detailed Solution1. Concentric circles have same center. Original center:

2g = -4 \implies g = -2, 2f = -6 \implies f = -3

. Center = (2, 3). 2. Circle passes through (-4, -5). Radius

r^2 = (2 - (-4))^2 + (3 - (-5))^2 = 6^2 + 8^2 = 100

. 3. New equation:

(x-2)^2 + (y-3)^2 = 100 \implies x^2 + y^2 - 4x - 6y - 87 = 0

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Class 11 Applied Mathematics

Circles &
Parabola

Exploring the beauty of conic sections. From the perfect symmetry of circles to the directional focus of parabolas.

The Circle

Symmetry in Motion

ℹ️

Standard Equation

A circle is the locus of a point which moves such that its distance from a fixed point (center) is constant (radius).\n\n

(x-h)^2 + (y-k)^2 = r^2

\n\nAt origin

(0,0)

x^2 + y^2 = r^2

Diameter & General Form

Diameter Form

(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0

Center Calculation

For $x^2 + y^2 + 2gx + 2fy + c = 0$:

Center:

(-g, -f)

, Radius:

\sqrt{g^2 + f^2 - c}

Circle Explorer

Visualize standard and general forms of a circle.

C(0, 0)

r = 5

Scale: 1 unit = 20px

Center h: 0

Center k: 0

Radius r: 5

Standard Form

(x - 0)^2 + (y - 0)^2 = 5^2

General Form ($2gx, 2fy$)

x^2 + y^2 +0x +0y -25.0 = 0

g = 0, f = 0

c = -25.0

Existence ConditionFor a real circle,

g^2 + f^2 - c > 0

. Current:

0^2 + 0^2 - (-25.0) = 25

(Valid)

Wait! Radius Check

The quantity $g^2 + f^2 - c$ determines the nature of the circle:

> 0

Real Circle

= 0

Point Circle

< 0

Imaginary

Tip: g = half of coeff of x, f = half of coeff of y

The Parabola

Parabola Lab

Explore standard forms and their geometric properties.

Focus(2, 0)

Directrixx = -2

Parameter (a): 2

AxisX-axis

LR Length4a = 8

Active Equation

y^2 = 4(2)x

Standard DefinitionDistance from Point to Focus = Distance from Point to Directrix.

Standard Forms Comparison

Four orientations of the standard parabola with vertex at $(0,0)$.

Equation	Focus	Directrix	Latus Rectum	Axis
$y^2 = 4ax$	$(a, 0)$	$x = -a$	$4a$	x-axis
$y^2 = -4ax$	$(-a, 0)$	$x = a$	$4a$	x-axis
$x^2 = 4ay$	$(0, a)$	$y = -a$	$4a$	y-axis
$x^2 = -4ay$	$(0, -a)$	$y = a$	$4a$	y-axis

Practice Laboratory

Conics Workouts

8 Important Problems

CBSE Exam Type

Concept Master

Find the equation of a circle whose center is (2, -3) and radius is 5.

✓ Show Solution

1. Given:

(h, k) = (2, -3), r = 5

2. Formula:

(x-h)^2 + (y-k)^2 = r^2

3. Substituting:

(x-2)^2 + (y+3)^2 = 25

4. Expanding:

x^2 + y^2 - 4x + 6y + 4 + 9 = 25

5. Result:

x^2 + y^2 - 4x + 6y - 12 = 0

EASY

Find the center and radius of the circle

x^2 + y^2 - 8x + 10y - 12 = 0

✓ Show Solution

1. General form:

x^2 + y^2 + 2gx + 2fy + c = 0

2. Comparing:

2g = -8 \implies g = -4

2f = 10 \implies f = 5

c = -12

3. Center:

(-g, -f) = (4, -5)

4. Radius:

r = \sqrt{g^2 + f^2 - c} = \sqrt{(-4)^2 + 5^2 - (-12)} = \sqrt{16 + 25 + 12} = \sqrt{53}

units

EASY

Find the equation of the circle passing through the points (3, 4) and (2, 1) as endpoints of a diameter.

✓ Show Solution

1. Diameter Form:

(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0

2. Substitute

(3, 4)

and

(2, 1)

(x-3)(x-2) + (y-4)(y-1) = 0

x^2 - 5x + 6 + y^2 - 5y + 4 = 0

5. Final:

x^2 + y^2 - 5x - 5y + 10 = 0

MEDIUM

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

✓ Show Solution

1. Let center be

(h, k)

. Radius

r^2 = (4-h)^2 + (1-k)^2 = (6-h)^2 + (5-k)^2

2. Expand & Solve for

h, k

16-8h+h^2+1-2k+k^2 = 36-12h+h^2+25-10k+k^2

4h + 8k = 44 \implies h + 2k = 11

4. Given center on

4x + y = 16 \implies 4h + k = 16

5. Solving simultaneously:

h=3, k=4

. Radius

r^2 = (4-3)^2 + (1-4)^2 = 1 + 9 = 10

6. Equation:

(x-3)^2 + (y-4)^2 = 10 \implies x^2 + y^2 - 6x - 8y + 15 = 0

HARD

Find the focus, axis and the length of the latus rectum for

y^2 = 12x

✓ Show Solution

1. Compare with

y^2 = 4ax \implies 4a = 12 \implies a = 3

2. Focus:

(a, 0) = (3, 0)

3. Axis: x-axis (

y=0

)

4. Length of Latus Rectum:

4a = 12

units

5. Directrix:

x = -3

EASY

Find the equation of the parabola whose focus is (0, -3) and directrix is y = 3.

✓ Show Solution

1. Focus is on y-axis, below origin. This is of the form

x^2 = -4ay

a = 3

(distance from vertex (0,0) to focus (0,-3)).

3. Equation:

x^2 = -4(3)y

4. Result:

x^2 = -12y

MEDIUM

Find the equation of the parabola with vertex at origin, axis along x-axis and passing through (2, 3).

✓ Show Solution

1. Axis along x-axis

\implies y^2 = 4ax

2. Passes through (2, 3)

\implies 3^2 = 4a(2) \implies 9 = 8a \implies a = 9/8

3. Equation:

y^2 = 4(9/8)x \implies y^2 = 4.5x

2y^2 = 9x

MEDIUM

Find the area of the triangle formed by the lines joining the vertex of the parabola

x^2=12y

to the ends of its latus rectum.

✓ Show Solution

1. Parabola

x^2 = 12y \implies 4a = 12 \implies a = 3

2. Vertex

O(0, 0)

. Focus

S(0, 3)

3. Latus Rectum ends:

x^2 = 4ay \implies x = \pm 2a = \pm 6

. Points are

(-6, 3)

and

(6, 3)

4. Base of triangle = distance between

(-6, 3)

and

(6, 3) = 12

units.

5. Height of triangle = distance from

O(0,0)

to focal line

y=3

3

units.

6. Area =

\frac{1}{2} \times 12 \times 3 = 18

sq. units.

HARD

Competitive Corner

Board & Previous Year Questions

CBSE 2023

Find the equation of the circle which touches both the axes and whose radius is 4 units.

Detailed Solution1. Since it touches both axes, center is

(\pm 4, \pm 4)

. 2. There are 4 such circles. Case 1 (Quadrant I): center

(4, 4), r=4

3. Equation:

(x-4)^2 + (y-4)^2 = 16 \implies x^2 + y^2 - 8x - 8y + 16 = 0

4. In general:

x^2 + y^2 \pm 8x \pm 8y + 16 = 0

CBSE 2021

Find the equation of the parabola whose vertex is (0,0) and focus is (-2, 0). Also find directrix.

Detailed Solution1. Vertex (0,0), Focus (-2, 0)

\implies

Opens left

\implies y^2 = -4ax

. 2.

a = 2

. 3. Equation:

y^2 = -8x

. 4. Directrix:

x = a \implies x = 2

Common Board Problem

Find the coordinates of the focus and the length of the latus rectum for the parabola $3x^2 = -8y$.

Detailed Solution1. Rewrite:

x^2 = -\frac{8}{3}y

2. Compare with

x^2 = -4ay \implies 4a = 8/3 \implies a = 2/3

3. Focus:

(0, -a) = (0, -2/3)

4. Latus Rectum length:

4a = 8/3

units.

Concept MCQ

If the equation $x^2 + y^2 + 2gx + 2fy + c = 0$ represents a circle, find the condition that the circle passes through the origin.

Detailed Solution1. Circle passes through

(0, 0)

. 2. Substitute

x=0, y=0

in the equation:

0^2 + 0^2 + 2g(0) + 2f(0) + c = 0

3. Result:

c = 0

Most Repeated

Find the equation of the circle which is concentric with $x^2+y^2-4x-6y-9=0$ and passes through ( -4, -5).

Detailed Solution1. Concentric circles have same center. Original center:

2g = -4 \implies g = -2, 2f = -6 \implies f = -3

. Center = (2, 3). 2. Circle passes through (-4, -5). Radius

r^2 = (2 - (-4))^2 + (3 - (-5))^2 = 6^2 + 8^2 = 100

. 3. New equation:

(x-2)^2 + (y-3)^2 = 100 \implies x^2 + y^2 - 4x - 6y - 87 = 0