1

Equation of a circle on some special cases

Janak Singh Saud [Mathematics Teacher]

Monday, May 1, 2023 Compiled By: Janak Singh saud

Balkumari, Koteshwor, Kathamandu

Equation of a circle on Some Special Cases

1. When a circle Touches X-axis

2. When a circle touches Y-axis

3. When a circle touches both the axes

4. When a circle passes through origin

5. When the circle touches the give line ax + by + c = 0

Monday, May 1, 2023Compiled By: Janak Singh saud

2

3

CONTINUE……….

Monday, May 1, 2023

Compiled By: Janak Singh saud

Case I: When a circle Touches X-axis

Case II: When a circle touches Y-axis

Case II: When a circle touches both the axes

Case IV :When a circle passes through origin

Case V: When the circle touches the give line ax + by + c = 0

4

Case I : When a circle Touches X-axis

When circle touches positive X-axis

Let, Center = (h, k). Then Radius = k units i.e. r = |

k|

So, equation of the circle is (x – h)2 + (y – k)2 = k2

Monday, May 1, 2023

Compiled By: Janak Singh saud

P(x, y)

C(h, k)

r

h A(h,0)

It is not a radius

k

Example:

Since, the circle with center (2, 4) touches X-axis

Center of the circle (h, k) = (2, 4)Radius ( r) = 4 units = kThen, using formula

(x – h)2 + (y- k)2 = r2 or, (x – 2)2 + (y – 4)2 = 42 or, x2 - 4x + 4 + y2 - 8y + 16 = 16or, x2 + y2 - 4x – 8y + 4 = 0Which is the required equation of the circle

Monday, May 1, 2023

5

Compiled By: Janak Singh saud

Find the equation of a circle with center (2, 4) which touches the X-axis

C(2,4)

2

r4

Monday, M

ay 1, 2023

6

Compiled By: Janak Singh saud

Example: Find the equation of a circle whose center is (- 4, - 3) and touches the X-axis.

C(-4,-3)

Since, the circle with center (2, 4) touches X-axis

Center of the circle (h, k) = (- 3, - 4)Radius ( r) = |- 3| = 3 units = k

Then, using formula (x – h)2 + (y- k)2 = r2 or, [x –( - 4)]2 + [y – (- 3)]2 = 42

or, (x + 4)2 + (y + 3)2 = 16

or, x2 + 8x + 16 + y2 + 6y + 9 = 16

or, x2 + y2 + 8x + 6y + 9 = 0Which is the required equation of the circle

r|3|

7

Case II :When a circle touches Y-axis

Monday, May 1, 2023Compiled By: Janak Singh Saud

P(x, y)

C(h, k)

h

h

k

A(0,

k)

When circle touches positive Y-axis

Let, Center = (h, k). Then

Radius = h units i.e. r = |h|

So, the equation of the circle is

(x- h)2 + (y – k)2 = h2

8

Example:

Since, the circle with center (2, 4) touches Y-axis

Center of the circle (h, k) = (- 3, - 4)Radius ( r) = |- 3| = 3 units = k

Then, using formula (x – h)2 + (y- k)2 = r2

or, [x –( - 3)]2 + [y – (- 4)]2 = 32

or, (x + 3)2 + (y + 4)2 = 9

or, x2 + 6x + 9 + y2 + 8y + 16 = 9

or, x2 + y2 +6x + 8y + 16 = 0Which is the required equation of the circle

Monday, May 1, 2023Compiled By: Janak Singh saud

Find the equation of a circle whose center is (- 3, - 4) and touches the Y-axis.

C(-3,-4

)

r|-3|

It is not a radius

9

Case III :When a circle touches both the axes

Monday, May 1, 2023Compiled By: Janak Singh saud

P(x, y)

C(h, k)

k

h

h A(h, 0)

B(0,

k)

Let, Center = (h, k). Then Radius = h or k units i.e. r = |h| = |k|So, the equation of the circle is

(x – h)2 + (y – k)2 = h2

or, (x – h)2 + (y – k)2 = k2

or, (x –h)2 + (y – h)2 = h2

or, (x – k)2 + (y – k)2 = k2

or, (x – r)2 + (y – k)2 = r2

When the circle touches both the positive axes

10

Example:

Here, the radius of a circle , r = 7 unitsSince, the circle touches both the positive axesSo, its center (h, k)=(r, r) = (7, 7)Now, using formula

(x – h)2 + (y- k)2 = r2 or, (x – 7)2 + (y- 7)2 = 72

or,x2 – 14x + 49 + y2 – 14y + 49 = 49

or, x2 + y2 - 14x – 14y + 49 = 0Which is the required equation of a circle.

Monday, May 1, 2023Compiled By: Janak Singh saud

Find the equation of the circle having radius 7 units and touches both the positive axes.

We know, in this case, r = h = k

r

r7

7

(7, 7)

11

Case IV : When a circle passes through origin

Monday, May 1, 2023

Compiled By: Janak Singh saud

When the circle with center (h, k) and radius r units passes through the origin O(0,0), the center of the circle will be

Center = (h, k)

Radius (r) =

i.e. r2 = h2 + k2

So, the equation of the circle is (x – h)2 + (y + k)2= h 2 + k2

C(h, k)h

kr

O(0,0)

By Pythagoras’ theorem for triangle OCAr2 = h2 + k2

k

A

NEXT WAYLet , the circle passes

through the origin, cuts x-as-s at A and y-axis at B such that OA = a and OB = b

AB is a diameter because the inscribed angle at origin is 900 (angle in semi-circle)

Now, using the equation of a circle in diameter form

Monday, M

ay 1, 2023

12

Compiled By: Janak Singh saud

B(0,

b)A(a, 0)

a

b

(x – x1)(x – x2 )+ (y–y1)(y– y2 )= 0

13

A circle passes through the origin O and making the intercepts 6 units and 7 units on the X-axis and Y-axis respectively.

So, it cuts the x-axis at A(6,0) and the y-axis at B(0,7)Since, the inscribed angle at O is 900

, AB is a diameter.Now, the equation of the circle in the

diameter form is given by(x – x1)(x – x2 )+ (y–y1)(y– y2 )= 0

or,(x – 6)(x – 0) + (y – 0 )(y – 7) = 0or, (x – 6) x + y (y – 7) = 0or, x2 - 6x + y2 - 7y = 0or, x2 + y2 - 6x – 7y = 0

Monday, May 1, 2023

Compiled By: Janak Singh saud

Example: Find the equation of the circle passing through the origin and making the intercepts of 6 units and 7 units on the positive X-axis and positive Y-axis respectively

B(0,

7)

A(6,0)

c

6

7

14

C(h, k)r

A

ax + by + c = 0

If the circle with center (h, k) and radius r units touches the straight line ax + by + c = 0 at a point A as shown in the adjoining figure, then the line segment CA is perpendicular to the given line, which is radius of the circle. Now, r = AC =

Hence, the equation of the circle becomes(x – h)2 + (y – k)2 =

Case V: When the circle touches the give line

ax + by + c = 0

Monday, May 1, 2023

Compiled By: Janak Singh saud

Monday, May 1, 2023

15

Compiled By: Janak Singh saud

Example: Find the equation of the circle with center at (6, -3 ) and touches the line x + 2y – 10 =0.

x + 2y – 10 =0.

C(h, k)

r

C(6,- 3)

Here, center of a circle (h, k) = (6, - 3)

Since, the circle touches the line x + 2y – 10 = 0

We have, radius = perpendicular distance from point (6, - 3) to the given line x + 2y – 10 = 0

Therefore, radius, r =

Now, using formula,(x – h)2 + (y- k)2 = r2 or, (x – 6)2 + [y-(- 3)]2 =

or, x2 – 12x + 36 + y2 +6y + 9 = 20or, x2 + y2 - 12x + 6y + 25 = 0Which is the required equation of the circle

16

Concentric Circles Monday, M

ay 1, 2023Com

piled By: Janak Singh saud

C1

C2

rR

C(-g, -f)

Circles having same center are called concentric

circles.

Circes C1 and C1 are concentric because center is same but radii different

17

THANK YOU

Monday, M

ay 1, 2023Com

piled By: Janak Singh saud